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3 years ago

If you kick a ball, you apply an impulse. The impulse is equal to

Physics

1 answer:

Murrr4er [49]3 years ago

6 0

Answer:

The correct answer is b. The change in momentum of the ball.

Explanation:

When you apply a force on an object with a direction in a period of time, you are changing its amount of movement (momentum).

We know that force (F) is equal to the mass (m) multiplied by the acceleration (a):

F= m x a

If we multiply both members of the equation by the time (t), we obtain:

F x t = m x a x t

And we know that a x t = velocity (v), so:

F x t = m x v

F x t= impulse

m x v = momentum

So, the impulse is equal to the change in momentum of the object.

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In the formula for magnetic deflection, FM is equal to qvB. Match each symbol in this formula with what it represents.

slava [35]

Answer:

1. $F_m$ -Deflection force

2.q-Charge

3.v-Velocity of the charge

4.B-Perpendicular magnetic field strength

Explanation:

We are given that magnetic deflection

$F_m=qvB$

We have to match each symbol in formula with what it represents.

In this formula

$F_m=$ Deflection force

q=Charge of particle

v=Velocity of charge

B=Perpendicular magnetic field strength

Therefore,

1. $F_m$ -Deflection force

2.q-Charge

3.v-Velocity of the charge

4.B-Perpendicular magnetic field strength

8 0

3 years ago

Read 2 more answers

A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 59.0 lb/in2. If the pump is a cylinder of lengt

bagirrra123 [75]

Answer:

∴The air cannot be made to flow in with the given pump at the given conditions.

Explanation:

Given:

gauge pressure of bicycle tyre, $P_g=59\ lb.in^{-2}$

length of cylinder of the pump, $l=17.4\ in$

area of the the cylinder of the pump, $a= 3\ in^2$

we have the density of air at STP, $\rho=4.4256\times 10^{-5}\ lb.in^{-2}$

The piston must be pushed more than the pressure inside the tyre:

$P_g=\rho\times V\div a$

$59=4.4256\times 10^{-5}\times a\times h\div a$

$h=13.33\times 10^5\ in$

∴The air cannot be made to flow in with the given pump at the given conditions.

4 0

3 years ago

The position function of a particle is given by s=2t3−3t2−63t,t⩾0.s=2t3−3t2−63t,t⩾0. where ss is measured in meters and tt in se

Kisachek [45]

Answer:

$t=5.72\ s$

Explanation:

Given:

the displacement as the function of time:

$s=2t^3-3t^2-63t$

here time is in seconds and the displacement in meters.

Now we differentiate this eq. of displacement to get the equation of velocity:

$v=\frac{d}{dt}(s) \\v=6t^2-6t-63$

According to given the velocity is $99\ m.s^{-1}$ at some time:

$99=6t^2-6t-63$

$6t^2-6t-162=0$

$t=5.72\ s$ & is the only time for (t>=0) instances when the particle will have a velocity of $99\ m.s^{-1}$ but in the opposite direction.

8 0

4 years ago

A 185 g block is pressed against a spring of force constant 1.60 kN/m until the block compresses the spring 10.0 cm. The spring

denis-greek [22]

Answer:

d = 5.10 m

Explanation:

As we know that here on the plane of the inclined there is no frictional force

So in these cases we can say that total mechanical energy will always remains conserved

so here we can say that

spring potential energy = gravitational potential energy of the block

as we know from the formula

$\frac{1}{2}kx^2 = mgh$

now plug in the values in it

$\frac{1}{2}(1.60 \times 10^3)(0.10)^2 = (0.185)(9.81)h$

$8 = 1.81 h$

$h = 4.42 m$

now as we know that the angle of inclination is 60 degree and height raised is 4.42 m

so here maximum distance moved along the inclined plane will be

$\frac{h}{d} = sin60$

$d = \frac{h}{sin60}$

$d = \frac{4.42}{sin60} = 5.10 m$

6 0

3 years ago

When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular

OLga [1]

Answer:

The force is $1237500N$

Explanation:

We can apply the following expression

$pressure=\frac{Force}{Area}$

To find the area of rectangular tent wall:

$Area=length *width\\Area=2.00m*0.750m\\Area=1.5m^{2}$

Now to find the force:

$pressure=825,000pa$

$pressure=\frac{Force}{Area}$

$Force=pressure*Area\\Force=825000pa*1.5m^{2} \\Force=1237500N$

7 0

3 years ago