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miss Akunina [59]
2 years ago
15

If the time of impact in a collision is extended by 3 times, by hour much is

Physics
1 answer:
vodomira [7]2 years ago
4 0

Answer:

The force of the impact would be smaller

Explanation:                                                          Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

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2 years ago
A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc
Fantom [35]

Answer:

1. 7.5 m

2. towards west side

explanation:

I hope it will help you

3 0
2 years ago
What is the relationship between the poles of a magnet? which direction does the magnet field move?
Assoli18 [71]
A magnet has a South Pole and a North Pole. South Pole and South Pole can't connect to her other, same as North and North. The same poles push each other away.
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4 0
3 years ago
PLEASE HELP ME 45 POINTS
sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a (1)

Mass B

\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a (2)

Where:

T - Tension force in the cord, measured in newtons.

m_{A}, m_{B} - Masses of blocks A and B, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g

(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g

a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g

If we know that m_{A} = 5\,kg, m_{B} = 2\,kg and g = 9.807\,\frac{m}{s^{2}}, then the acceleration of the masses is:

a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)

a = 4.203\,\frac{m}{s^{2}}

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

T = m_{B}\cdot (a+g)

If we know that m_{B} = 2\,kg, g = 9.807\,\frac{m}{s^{2}} and a = 4.203\,\frac{m}{s^{2}}, then the tension force in the cord:

T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}  \right)

T = 28.02\,N

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

\Delta y  = \frac{1}{2}\cdot a\cdot t^{2} (3)

Where:

a - Net acceleration, measured in meters per square second.

t - Time, measured in seconds.

\Delta y - Covered distance, measured in meters.

If we know that a = 4.203\,\frac{m}{s^{2}} and \Delta y = 1.5\,m, then the time taken by the system is:

t = \sqrt{\frac{2\cdot \Delta y}{a} }

t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }

t \approx 0.845\,s

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

v = a\cdot t (4)

Where v is the final speed of the system, measured in meters per second.

If we know that a = 4.203\,\frac{m}{s^{2}} and t \approx 0.845\,s, then the final speed of the system is:

v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)

v = 3.551\,\frac{m}{s}

The final speed of the system is 3.551 meters per second.

8 0
2 years ago
help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help ​
kompoz [17]

Answer:

0.5 m/s².

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

a = (v – u) / t

a = (10 – 0) / 20

a = 10/20

a = 0.5 m/s²

Therefore, the acceleration of the car is 0.5 m/s².

6 0
2 years ago
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