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anyanavicka [17]
3 years ago
6

What is mass times acceleration?

Physics
1 answer:
frozen [14]3 years ago
6 0

Answer:

Force

Explanation:

The mass of an object is the quantity of matter it contains. It is measured in kilograms.

Acceleration is the ratio of the change in the velocity of an object to the change in time. It is measured in m/s^{2}.

When the mass of an object is multiplied with its acceleration, this gives the average force applied on the object. As force is defined as agent that can change the state of an object.

i.e     F = m × a

where F is the force, m is the mass of the object and a its acceleration.

The two major classes of force are; contact force and field force.

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The formula Sequals16tsquared is used to approximate the distance​ S, in​ feet, that an object falls freely from rest in t secon
Tom [10]

S = 16t² where t = time in seconds.

Let S = 1190ft then solve for t:

16t² = 1190

t² = 74.4

t = 8.62s

5 0
3 years ago
A 2.0 kg pendulum has an initial total energy of 20 J. Calculate the energy lost as heat if the pendulum is 0.10 m high and is t
Maurinko [17]

The correct answer is (A) 2.0 J

Total energy of the pendulum is the sum of its kinetic and potential energy. At the instant of time, when the pendulum is at a height <em>h</em> and has a speed <em>v, </em>Its energy is given by,

E=mgh+\frac{1}{2} mv^2

Substitute 2.0 kg for <em>m</em>, the mass of the pendulum, 9.81 m/s² for <em>g</em>, the acceleration due to gravity, 0.10 m for <em>h and 4.0 m/s for </em>v<em>.</em>

E=mgh+\frac{1}{2} mv^2\\ =(2.0kg)(9.81m/s^2)(0.10 m)+\frac{1}{2}(2.0kg)(4.0m/s)^2\\ =17.962J

The pendulum has an initial energy of 20 J. the energy lost is given by,

\Delta E=(20J)-(17.962J)\\ =2.038J=2.0J

Thus, the energy lost by the pendulum is (A) 2.0 J

4 0
3 years ago
A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi
Marrrta [24]

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, \mu = 0.19

Now,

We know that the circumference of the circle is given by:

C = 2\pi R

27 = 2\pi R

R = \frac{27}{2\pi } = 4.29\ in

Also, in case of rolling, we know that:

Angular velocity, \omega = \frac{v}{R}

v = \omega R

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

m\omega' = m\omega + I\omega

Moment of Inertia, I = \frac{2}{5}mR^{2} = mk^{2}

mv'R = mvR + mk^{2}\times \frac{v}{R}

v'R = vR + k^{2}\times \frac{v}{R}

276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}

1184.04 - 756.903 = 4.29v

v = 99.56 in/s

Now,

Friction force, f = \mu mg

Also, acceleration of the ball can be computed as:

\mu mg = - ma

a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}

Now, the distance, d covered by the ball before rolling without slipping:

v^{2} = v'^{2} + 2ad

99.56^{2} = 276^{2} + 2\times (- 73.357)d

99.56^{2} - 276^{2} = 2\times (- 73.357)d

d = 451.65 in = 37.53 ft

7 0
3 years ago
A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.75 m from the slits. If the bright
Mariulka [41]

Answer:

7.55\times 10^{-7} m

Explanation:

We are given that

d=0.23 mm=0.23\times 10^{-3} m

1mm=10^{-3} m

Screen is placed  from the slits at distance ,L=4.75 m

The bright interference fringes on the screen are separated  by 1.56 cm.

\Delta y=1.56 cm=1.56\times 10^{-2} m

1 m=100 cm

We have to find the wavelength of laser light.

We know that

\Delta y=\frac{\lambda L}{d}

Substitute the values

1.56\times 10^{-2}=\frac{\lambda\times 4.75}{0.23\times 10^{-3}}

\lambda=\frac{1.56\times 10^{-2}\times 0.23\times 10^{-3}}{4.75}

\lambda=7.55\times 10^{-7} m

4 0
3 years ago
Which best distinguishes between the inner and outer planets?
Murrr4er [49]
Either a or b 
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3 0
4 years ago
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