Force can change the ____________ and ______________

A ball of mass 4.5 kg moving with speed of 2.2 m/s in the +x-direction hits a wall and bounces back with the same speed in the x

Levart [38]

Answer:

The change of the momentum of the ball is $-19.8\, \frac{mkg}{s}$

Explanation:

We should find $\varDelta\overrightarrow{p}=\overrightarrow{p_{f}}-\overrightarrow{p_{i}}$ (1)with $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum. Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (1):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (2)

It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall $\overrightarrow{v_{i}}=+2.2\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-2.2\, \frac{m}{s}$ so (2) becomes:

$\varDelta\overrightarrow{p}=m(-2.2- (+2.2))=-(4.5)(4.4)$

$\varDelta\overrightarrow{p}=-19.8\, \frac{mkg}{s}$

8 0

3 years ago

A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

astraxan [27]

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$

7 0

3 years ago

The combination of all forces that act on an object are (2 points)

qaws [65]

Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

5 0

3 years ago

-2 5/8 is bigger than -5/12

Neko [114]

Answer:

hey...................................

8 0

3 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

Force can change the ____________ and _______________ of an object​

Force can change the and ___ of an object