Answer:
20 degrees
Explanation:
Given that,
Speeds of both snowballs is 34.8 m/s
The first one is thrown at an angle of 70◦ with respect to the horizontal. We need to find the angle at which should the second snowball be thrown to arrive at the same point as the first.
We need to find the angle at which the second snowball be thrown to arrive at the same point as the first. We can find the distances of both balls and equate them.
It means, range of both projectiles are equal. So,
\(\dfrac{u^2\sin2\theta_1}{g}=\dfrac{u^2\sin2\theta_2}{g}\)
We have, \(\theta_1=70^{\circ}\)
So,
\(\theta_2=\dfrac{1}{2}(\sin^{-1}(\sin2\theta_1))\\\\\theta_2=\dfrac{1}{2}(\sin^{-1}(\sin(140))\\\\\theta_2=20^{\circ}\)
Hence, the second snowball is thrown at an angle of 20 degrees.
A battery of internal resistance 1 Ω and emf of 12 volt connected to external resistance of 79 Ω.Calculate the power delivered by the battery to the circuit.
Answer:
P = 1.8 Watt
Explanation:
Electromotive Force (EMF)
It consists of a DC source ε with small internal resistance Ri. When it's connected to an external resistance Re a current i flows by the circuit whose value is given by:
\(\displaystyle i =\frac{\varepsilon }{R_i+R_e}\)
The internal resistance of the battery is Ri=1 ohm and its emf is ε=12 V. When connected to an external resistance of Re=79 ohm, the current is:
\(\displaystyle i =\frac{12}{1+79}=\frac{12}{80}\)
i = 0.15 A
The power delivered by the battery is:
P = ε.i
P = 12 V * 0.15 A
P = 1.8 Watt
what will it be the critical angle if light is incident from glass n=1.5 in to the air n=1
Answer:
The formula, which is fairly easy to derive, is:
The sine of the critical angle = refractive index of the “refracting medium” divided by refractive index of the “incident medium”.
In this case the “refractive medium” is air and the “incident medium” is glass.
The refractive index of air is about 1 (almost exactly).
So, in this case, the sine of the critical angle is 1 divided by 1.5 = 0.6667.
You will find the answer, from sine tables, to be approximately 41.81 degrees.
Remember that the refractive index of air is slightly more than 1. Using a better value for refractive index of air, I get the critical angle to be about
During the Apollo Moon missions, astronauts were concerned about the
effects of Newton's third law of motion when:
A. they detected incoming space debris, because the inertia of the
debris could damage their ship.
B. they reentered Earth's atmosphere, because they were carrying
heavy Moon rocks
C. They ejected waste water, because they were concerned that it would push them off course.
D. they took off from Earth, because a large force was required to generate large acceleration
Answer:
C
Explanation:
Option C is correct since it is the effect of Newton’s third law.
A hiker with a skin surface area of 1.3 m² is protected from hypothermia (the cold) by a close-fitting sleeping bag 30 mm thick. If her skin temperature is 34°C and she can safely lose 85 W of heat by conduction through the sleeping bag, what is the lowest outside temperature for which the sleeping bag provides adequate protection? Ignore heat losses due to convection or radiation. Coefficient of thermal conductivity of the sleeping bag = 0.019 Wm¹¹°C-¹
Answer: The lowest outside temperature for which the sleeping bag provides adequate protection is approximately 89.61°C below the hiker's skin temperature of 34°C.
Explanation:
To find the lowest outside temperature for which the sleeping bag provides adequate protection, we need to determine the rate of heat loss through conduction and compare it to the heat loss the hiker can safely tolerate.
The rate of heat loss through conduction can be calculated using the formula:
Q = (k * A * ΔT) / d
Where:
Q is the rate of heat transfer (in Watts)
k is the coefficient of thermal conductivity (0.019 Wm¹¹°C-¹ in this case)
A is the surface area (1.3 m² in this case)
ΔT is the temperature difference (in this case, the difference between the skin temperature and the outside temperature)
d is the thickness of the sleeping bag (30 mm, which needs to be converted to meters by dividing by 1000)
Let's plug in the values:
Q = (0.019 * 1.3 * ΔT) / (30 / 1000)
The hiker can safely lose 85 W of heat, so we can set up the equation:
85 = (0.019 * 1.3 * ΔT) / (30 / 1000)
To solve for ΔT, we can rearrange the equation:
ΔT = (85 * (30 / 1000)) / (0.019 * 1.3)
ΔT ≈ 89.61°C
The four particles as connected by rods of negligible mass as fig below. if the origin is the canter of rectangle and the system rotates in the XY plane about the Z axis with an rad angular speed of 12. calculate S a) The moment of inertia of the system about Z axis and b) The rotational kinetic energy of the system 3.00 kg 2.00 kg y(m) 2.00 kg 6.00 m 4.00 kg ---x(m)
The moment of inertia of the system about the Z-axis is 245 kg m², and the rotational kinetic energy of the system is 21168 J.
The moment of inertia of a system about its axis of rotation is the sum of the products of the masses of its constituents and the square of their respective distances from the axis of rotation.
The radius of the rectangular plate is 6 m, and the distance of each particle from the center is half of the sides of the rectangle, which are 4 m and 3 m.
Therefore, using the parallel axis theorem, we get the moment of inertia of the system about the Z-axis as shown below.
\(Iz = ICM + MR^{2}\)
(1)We can obtain the moment of inertia of the rectangle about its center as: \(ICM = (1/12) ML^{2}\)
(2) where M is the mass of the rectangle, and L is the length of the rectangle.
Substituting values, we get: ICM = \((1/12) $\times$ 3.00 $\times$ (4^{2} + 6^{2} )\)
ICM = \(5 kg m^{2}\)
Using the parallel axis theorem, the moment of inertia of the four particles about the center of the rectangle is:
\(IP = 4 $\times$ [(1/12) $\times$ 2.00 $\times$ (4^{2} + 3^{2})] + 2.00 $\times$ (3^{2}) + 4.00 $\times$ (4^{2})IP = 97 kg m^{2}\)
The moment of inertia of the system about Z-axis is: \(Iz = ICM + MR^{2} Iz = 5 kg m^{2} + 3.00 kg $\times$ (6^{2} ) + 4 $\times$ [(4^{2}+ 3^{2} )/4] Iz = 245 kg m^{2}\)
The kinetic energy of a rotating body is given as:\(K.E. = (1/2) I\omega^{2}\) where I is the moment of inertia of the system, and ω is the angular velocity of the system.
The rotational kinetic energy of the system is:\(K.E. = (1/2) I\omega^{2} K.E. = (1/2) $\times$ 245 $\times$ (12)^{2} K.E. = 21168 J\)
2)\(I\omega^{2} K.E. = (1/2) $\times$ 245 $\times$ (12)^{2} K.E. = 21168 J\)
Therefore, the moment of inertia of the system about the Z-axis is 245 kg m², and the rotational kinetic energy of the system is 21168 J.
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Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure to label each axis and note the units used in the measurements. Be sure to draw a smooth curve through the points. Do not just connect the dots. Upload your data and graph in the essay box below and answer the following questions. Did the car travel at a constant speed? What was the average speed of the car? What are some practical applications for determining the motion of an object?
Answer:
yes it was a constant speed and the car traveled 10 meters in 20 seconds.
Explanation:
Answer:
It's a constant speed and the car traveled 10 meters in 20 seconds.
hope this helps!
Explanation:
What is audience going to be in the nine grade physic MCAS 
The audience for the 9th grade physics MCAS is students in the 9th grade who are taking the Massachusetts Comprehensive Assessment System (MCAS) physics exam.
How to explain the informationThe exam is designed to assess students' knowledge and skills in physics, and it is used to make decisions about students' placement in courses, graduation requirements, and eligibility for college and career programs.
The 9th grade physics MCAS is a multiple-choice exam that consists of 50 questions. The questions are based on the Massachusetts Curriculum Framework for Science, which outlines the standards that students are expected to meet in physics.
Students who take the 9th grade physics MCAS must score at least proficient in order to meet the state's graduation requirements. Students who score below proficient may be required to take additional courses or remediation.
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The drivers of two cars having equal speeds hit their brakes at the same time, but car A has four times the acceleration as car B .
Part A
If car A travels a distance D before stopping, how far (in terms of D ) will car B go before stopping?
Express your answer as an integer.
Part B
If car B stops in time T , how long (in terms of T ) will it take for car A to stop?
Express your answer to two significant figures
Car A's time to halt, tA = 0.33T. This is the right response to the question that was asked. The answer has two significant figures.
What is a good illustration of two vehicles with similar speeds but differing velocities?Provide an illustration of two vehicles with similar speeds but differing velocities. Example: Two 40 km/hr automobiles, one heading east and the other north.
If two cars accelerate at the same rate, do they move at the same speed? If not, why not?Equal accelerations do not necessarily imply identical velocities, and vice versa. For instance, even if both of your cars have the same acceleration, if one of them accelerates first, it will certainly go more quickly than the other.
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Which of the following is not part of the control system? steering wheel exhaust pipe brake accelerator
Answer:
Brakes unevenly adjusted: Brakes pulling in one direction or the other can lead to a skid. Tires with worn tread: Tread is necessary for traction in wet weather
Explanation:
Answer: exhaust pipe
Explanation: It doesn't really control anything so I'm assuming this is the answer
How did earth change about 2.5 billion years ago when many organisms began using photosynthesis to make food
A. The amount of oxygen in the atmosphere increased
B. Mass extinctions occurred
C. The oceans became larger
D. Rainfall increased
A bat emits a sonar sound wave
(343 m/s) that bounces off a
mosquito 8.42 m away. How
much time elapses between
when the bat emits the sound
and when it hears the echo?
(Unit = s)
Please help I am very confused with this topic
The time between when the bat emits the sound and when it hears the echo is 0.05 s
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 sThus, the time between when the bat emits the sound and when it hears the echo is 0.05 s
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Answer:
0.0491
Explanation:
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
An organ pipe of length L has one end closed but the other end open. What is the wavelength of the fundamental node emitted?
a. Slightly smaller than 4 L
b. Slightly larger than 4 L c. Roughly equal to 3/2
d. Slightly larger than 2 L
Answer:analize a afirmacao a seguir e tudo que envolve o gerenciamento da marca e que ultrapassa as acoes com objetivos economicos e refere se a cultura principios e valores
Explanation:
What is the difference between center of mass and center of gravity?
Answer:
Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.
Answer:
Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field.
Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.
hope this helps you ☺️☺️
what would happen to the gravitational force between earth and the moon if the distance between them increased?
Answer:
The gravitation force would weaken.
Explanation:
It would weaken as they would have less of a effect on eachother. Additionally, the sun would most likely grab the moon and catapult it.. but we dont talk about it.
An object moves at constant speed in a circle. Which of the following is true:
A. A net force in the direction of motion acts on the object.
B. A net force pointing away from the center of the circle acts on the object.
C. A net force pointing towards the center of the circle acts on the object.
D. The net force acting on the object is zero.
AC power switches direction at a frequency of 60 Hz. What is the period of this oscillation?
When the direction of a current changes 60 times in a second, the current is said to be 60 Hz.
An AC power grid with a 60 Hz frequency has a period, what is it?While the passing current does not affect the wavelength, the operating frequency does.A sine wave's period, which is defined as the interval between two identical spots on the wave, will be roughly 16.6 milliseconds at a rate of 60Hz (60 oscillations per second), as depicted in the diagram.
Describe a 60 Hz AC?A frequent word for 60 cycles per second, or more commonly 60 Hertz, is the frequency of dc power (Hz).In honor of German physicist Heinrich Hertz, who established the transmission and presence of electric oscillations, the latter unit is used.
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Statement: Baseball is not a dangerous sport unless either a bat breaks or a player is hit with a ball.
Key: B = A bat breaks.
D = Baseball is a dangerous sport.
P = A player is hit with a ball.
Answer:
P
Explanation:
If the bat break it will get replaced and you will get another bat,but if u get hit with the ball it's dangerous in many ways
How many meters are there in 20kilometers?
There are 20,000 meters in 20 kilometers. , we can represent this as:
$20 \, \text{km} = 20,000 \, \text{m}$
\(\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}\)
♥️ \(\large{\textcolor{red}{\underline{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}\)
Answer:20000 metres i think
Explanation:
I KM should have 1000m please vote for brainliest
Emory pushes the box with 20 Newtons of force. If the box is 4kg, how fast will the box accelerate?
A ball accelerates at a rate of 5 m/s2 when a 25 N force is applied to it. If the same
amount of force is applied to a 7 kg ball, what will happen to the ball's acceleration?
The acceleration of the ball would decrease when the same amount of force is applied.
Given the following data:
Acceleration = 5 \(m/s^2\)Force = 25 NewtonMass = 7 kgTo determine the new acceleration of the ball when the same amount of force is applied, we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
\(Acceleration = \frac{Force}{Mass}\)
Substituting the given parameters into the formula, we have;
\(Acceleration = \frac{25}{7}\)
Acceleration = 3.57 \(m/s^2\)
Therefore, the acceleration of the ball would decrease when the same amount of force is applied.
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A generator is a machine which makes:
electricity
magnetism
magnetic materials
difference between rest and motion
100 g of water at 25 °C is poured into an insulating cup. 50 g of ice at 0 °C is added to the water. The water is stirred until the temperature of the water has fallen to 0 °C. 18 g of ice remains unmelted. The specific heat capacity of water is 4.2 J / g °C. Which value does this experiment give for the specific latent heat of fusion of ice?
The specific latent heat of fusion of ice obtained from this experiment is approximately 583.33 J/g.
To determine the specific latent heat of fusion of ice using the given experiment, we need to consider the energy transferred during the process.First, we need to calculate the energy lost by the water to cool down from 25 °C to 0 °C. The energy lost is given by:
Q1 = m1 * c * ΔT1
Where:
m1 = mass of water = 100 g
c = specific heat capacity of water = 4.2 J/g °C
ΔT1 = change in temperature = (0 °C - 25 °C) = -25 °C
Q1 = 100 g * 4.2 J/g °C * (-25 °C) = -10,500 J
Next, we calculate the energy released by the water to freeze and cool the remaining ice. The energy released is given by:
Q2 = m2 * Lf
Where:
m2 = mass of ice = 18 g
Lf = specific latent heat of fusion of ice (to be determined)
Q2 = 18 g * Lf
Since energy is conserved in the system, the energy lost by the water (Q1) is equal to the energy released by the water (Q2):
-10,500 J = 18 g * Lf
Solving for Lf:
Lf = -10,500 J / 18 g = -583.33 J/g
The negative sign indicates that energy is being released during the process of freezing.
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Technician A says that cabin filters are accessible behind the glove compartment. Technician B says that cabin filters are accessible from under the hood. Who is correct? A. Technician B only B. Both Technicians A and B C. Technician A only D. Neither Technician A nor B
Answer:
A
but it depends on the car
Explanation:
Problem
THE FLIGHT OF A BALL A ball is launched at 5.5 m/s at 76° above
the horizontal. It starts and lands at the same distance from the
ground. What are the maximum height above its launch level and the
flight time of the ball?
1. The maximum height above its launch level is 1.45 m
2. The time of flight of the ball is 1.1 s
1. How do I determine the maximum height?
From the question given above, the following data were obtained:
Initial velocity (u) = 5.5 m/sAngle of projection (θ) = 76 °Acceleration due to gravity (g) = 9.8 m/s²Maximum height (H) =?The maximum height can be obatianed as follow:
H = u²Sine²θ / 2g
H = [5.5² × (Sine 76)²] / (2 × 9.8)
Maximum height = 1.45 m
How do I determine the time of flight?
The time of flight of the ball can be obtained as follow:
Initial velocity (u) = 5.5 m/sAngle of projection (θ) = 76 °Acceleration due to gravity (g) = 9.8 m/s²Time of flight (T) = ?T = 2uSineθ / g
T = [2 × 5.5 × Sine 76] / 9.8
Time of flight = 1.1 s
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Describe the relationship between frequency and wavelength of the visible light spectrum?
Select True or False to describe each statement about the forces between the cat and th
Gravity is a contact force on both the cat and the flap.
The push from the flap is a contact force on the cat...
The push from the cat is a contact force on the flap.
Search
hn
True
2 3
False
4
5
6
Answer:
Gravity is a contact force on both the cat and the flap. - False
The push from the flap is a contact force on the cat. - True
The push from the cat is a contact force on the flap. - True
Explanation:
Gravity is a contact force on both the cat and the flap. - False
Gravity is a non-contact force that acts between objects with mass. It does not require direct contact between the objects.
The push from the flap is a contact force on the cat. - True
When the flap pushes against the cat, it exerts a contact force on the cat.
The push from the cat is a contact force on the flap. - True
When the cat pushes against the flap, it exerts a contact force on the flap.
g The system is released from rest with no slack in the cable and with the spring unstretched. Determine the distanced s traveled by the 30-kg cart before it comes to rest (a) if m approaches zero and (b) if m
Solution :
It is given that a system is released from rest where no slack is attached to the cable and the spring is unstretched.
Mass of the cart = 30 kg
Applying the conservation of energy of the system,
Loss of the gravitational potential energy (PE) = the energy stored in the spring.
i.e., \(mg \Delta h = \frac{1}{2}kx^2\)
\(mg (x\sin 25^\circ)=\frac{1}{2}kx^2\)
\((30 \times 9.81) (x\sin 25^\circ)=\frac{1}{2}(125)x^2\)
x = 1.98
Therefore the distance travelled by the 30 kg cart is 1.98 m.
find the acceloration for xf=2.5m+(8.7m/s)t+(2.9m/s2)t2
When i put in 2.9 m/s2 it said i was wrong
Answer:
5.8 m/s²
Explanation:
The kinematic equation relating displacement, initial velocity, acceleration, and time, is:
Δx = v₀ t + ½ at²
So the coefficient of the t² term is ½ a.
½ a = 2.9 m/s²
a = 5.8 m/s²