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3 years ago

Select the correct answer.

Physics

1 answer:

Kipish [7]3 years ago

4 0

Answer:

ОА.

The net force acting on the car from all directions is zero.

Explanation:

Constant speed along a straight line

It means constant magnitude of velocity without chainging direction

It means constant velocity. Which means acclecration is zero

By Newton's second law:

$F=ma$

Net force is also zero

You might be interested in

An instrument for observing very small objects is

VladimirAG [237]

Microscope
A microscope shines a light through a slide with the sample of the object on and you look through a lens that allows you to magnify it by a very significant amount depending on what kind of microscope you have (a powerful electron microscope can magnify to the atomic level).

6 0

4 years ago

if 130N centripetal force is needed to keep a 0.45kg ball that is attached to a string that is 0.7m long to complete 5 full rota

andrezito [222]

Answer:

centripetal acceleration of the ball is 6.9 m/s/s

tangential speed of the ball is 2.2 m/s

Explanation:

As we know that ball complete 5 rotations in 10 seconds

so frequency of rotation of ball is given as

$f = \frac{5}{10} = 0.5 Hz$

now we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2(\pi)(0.5)$

$\omega = \pi rad/s$

Now centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \pi^2 (0.7)$

$a_c = 6.9 m/s^2$

now the velocity of the ball at this angular frequency is given as

$v = R\omega$

$v = 0.7 (\pi)$

$v = 2.2 m/s$

4 0

3 years ago

What is the input of energy for a device that provides 50 units of useful energy and has an efficiency of 0.55?

ki77a [65]

If that's the case, then

      50 units = 0.55 x the input energy

Divide each side by 0.55 :

   50 units/0.55 = the input energy =

   <span> 90 and 10/11 units</span>

3 0

3 years ago

Please help!! giving a lot of points

Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

=》

$67500 = 4500 \times 10 \times h$

=》

$67500 = 45000 \times h$

=》

$h = \dfrac{67500}{45000}$

=》

$h = 1.5 \: m$

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

mass = 4500 kg
kinetic energy = 63000 j

we know,

=》

$k.e = \dfrac{1}{2} mv {}^{2}$

=》

$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

=》

${v}^{2} = \dfrac{63000 \times 2}{4500}$

=》

${v}^{2} = 28$

=》

$v = \sqrt{28}$

=》

$v = 2 \sqrt{7} \: \: ms {}^{ - 1}$

=》

$5.29 \: \: ms {}^{ - 1}$

7 0

3 years ago

A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir

elena55 [62]

In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.

Let us list out what we know from the question so that we can decide on the equation to be used.

We know that Initial Y Velocity $V_{iy}$ = 8.4 m/s

Acceleration in the Y direction $a_{y}$ = -9.8 m/ $s^{2}$ , since the acceleration due to gravity points in the downward direction.

Final Y Velocity $V_{fy}$ = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.

Time taken t = ?

From the list above, it is easy to see that the equation that best suits our purpose here is $V_{fy} = V_{iy} + a_{y}t$

Plugging in the numbers, we get 0 = 8.4 - (9.8)t

Solving for t, we get t = 0.857 s

Therefore, the snowball takes 0.86 seconds to reach its highest point.

8 0

4 years ago