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3 years ago

Infer the direction of the net force acting on a car as it slows down and turns right.

2 answers:

6 0

Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.

xxTIMURxx [149]3 years ago

3 0

Explanation:

From Newton's second law of motion:

Force acting on a body, F = ma

where, m is the mass and a is the acceleration.

When the car is slowing down, it is actually decelerating. The direction of force is the direction of the acceleration. Since, the car is decelerating, the direction of net force acting would be opposite to the direction of motion of car. And when the car turns right, the net force is also directed towards right.

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State and explain guass law?

julia-pushkina [17]

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3 years ago

While in empty space, an astronaut throws a ball at a velocity of 11 m/s. what will the velocity of the ball be after it has tra

Kisachek [45]

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3 years ago

In a computer, the __________________________ chips are magnetic.

Sphinxa [80]

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3 years ago

a 3.0 kg mass moving to the right at 1.4 m/s collides in a perfectly inelastic collision with a 2.0 kg mass initially at rest. w

Sedbober [7]

The velocity of the combined mass after the collision is 0.84 ms-1.

Explanation:

According to law of conservation of momentum, the change in momentum before collision will be equal to the change in momentum of the objects after collision in isolated system.

But as it is perfectly inelastic collision in the present case, the final momentum will be based on the product of total mass of both the object with the velocity with which the collision occurred. This form is attained from the law of conservation of momentum as shown below:

So as law of conservation of momentum,

$M_{1} U_{1}+M_{2} U_{2}=M_{1} V_{1}+M_{2} V_{2}$

Here $M_{1}$ = 3 kg and $M_{2}$ = 2 kg are the masses of objects 1 and 2, $U_{1}$ = 1.4 m/s and $U_{2}$ = 0 are the initial velocities of object 1 and object 2, $V_{1}$ and $V_{2}$ are the final velocities of the objects.

So after collision, object 1 get sticked to object 2 and move together with equal velocity $V_{1}$ = $V_{2}$ = $V_{f}$ . Thus the above equation will become,

$M_{1} U_{1}+M_{2} U_{2}=\left(M_{1}+M_{2}\right) V_{f}$

So the final velocity is

$V_{f}=\frac{M_{1} U_{1}+M_{2} U_{2}}{\left(M_{1}+M_{2}\right)}$

Thus,

$V_{f}=\frac{(3 \times 1.4+2 \times 0)}{(3+2)}=\frac{4.2}{5}$ = 0.84 ms-1.

8 0

3 years ago

If the time from stimulus to response from the distal stimulating electrodes is 9.5ms and the time from stimulus to response fro

S_A_V [24]

Answer:

55m/s

Explanation:

conduction velocity (meter/second) is of a nerve is calculated by measuring the distance (millimeters) between two stimulation sites and dividing by the difference in latency (milliseconds) from the proximal.

mathematically is given as:

Conduction Velocity (meters/second) ={Distance (millimeters)}/{Proximal Latency- Distal Latency (millisecond)

C.V= (55mm)/ (10.5-9.5)ms

=55m/s

3 0

3 years ago

Read 2 more answers