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2 years ago

A tow truck exerts a force of 3000 n on a car, accelerating it at 2 meters per second per second. what is the mass of the car?

1 answer:

6 0

If the force is 3,000 N and the acceleration is 2 m/s/s. The mass of the car is 1,500 grams (or whatever unit of mass or weight is used in the answer- pounds, kilograms e.t.c.).

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Which sentence in the passage describes retrograde motion of a planet

Soloha48 [4]

Answer:

what is the passage? I don’t see it

Explanation:

4 0

3 years ago

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti

ivanzaharov [21]

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

7 0

2 years ago

3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a

inessss [21]

Answer:

The average angular acceleration is $\alpha =125.487 rad /s^2$

Explanation:

From the question we are told that

The length of the bat is $l = 0.85m$ \

The initial linear velocity is $u = 0 m/s$

The time is $t = 0.15s$

The velocity at t is $v = 16 m/s$

Generally average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_o}{t}$

Where $w_f$ is the finial angular velocity which is mathematically evaluated as

$w_f = \frac{v}{l}$

$w_f = \frac{16}{0.85}$

$= 18.823 rad/s$

and $w_o$ is the initial angular velocity which is zero since initial linear velocity is zero

$\alpha = \frac{18.823 - 0}{0.15}$

$\alpha =125.487 rad /s^2$

5 0

3 years ago

Who are the scientists given credit for discovering electrons, protons, and neutrons?

nydimaria [60]

J.J. Thompson is the scientist who recieved credit for discovering them.

7 0

2 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$