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3 years ago

What do we call a change in the car’s speed or direction? Help!!

Physics

2 answers:

zhuklara [117]3 years ago

8 0

The answer for this question is acceleration

worty [1.4K]3 years ago

6 0

When a car has a change of speed it is accelerating

Brainiest plz

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A boat of mass 225 kg drifts along a river at a speed of 21 m/s to the west. what impulse is required to decrease the speed of t

zzz [600]

The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
$J=\Delta p=m \Delta v$
where
m=225 kg is the mass of the boat
$\Delta v=v_f-v_i=15 m/s-21 m/s=-6 m/s$ is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
$J=m\Delta v=(225 kg)(-6 m/s)=-1350 N s$
and the negative sign means the direction of the impulse is against the direction of motion of the boat.

8 0

3 years ago

Read 2 more answers

A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame

Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

$f=\dfrac{v}{2l}$

v is the speed of the mass

Speed is given by :

$v=\sqrt{\dfrac{T}{\mu}}$

$\mu$ is the mass per unit length, $\mu=\dfrac{m}{l}$

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

T is the tension in the wire,

$f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

$T=4f^2lm$

$T=4(120)^2\times 0.75\times 0.012$

T = 518.4 N

Tension in the wire, T = m' g

$m'=\dfrac{T}{g}$

$m'=\dfrac{518.4}{9.8}$

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

6 0

3 years ago

engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi

Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

where ;

$m_1=mass of object 1\\v_1i=initial velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2$

Given in the question that;

$m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?$

Apply the formula as;

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

<u>57 m/s = v₂f { nearest whole number}</u>

4 0

2 years ago

Read 2 more answers

Question 15 of 15

castortr0y [4]

Answer:

a and b

Explanation:

are the correct answers

7 0

3 years ago

Please view image attached!

Marat540 [252]

Answer:

you would expect a change in the graph

Explanation:

8 0

2 years ago