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3 years ago

A 15kg ball accelerates at a rate of 3m/s/s. What force was required?

Physics

1 answer:

hoa [83]3 years ago

3 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15 × 3

We have the final answer as

Hope this helps you

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What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds

Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

$F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N$

The net power needed to change the speed of the vehicle is calculated as;

$P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W$

3 0

3 years ago

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu

seropon [69]

Answer:

$\theta=50\ revolution$

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, $\omega_i=600\ rpm=62.83\ rad/s$

Finally it shuts off, $\omega_f=0$

Time taken, t = 10 s

Let $\alpha$ is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{0-62.83}{10}$

$\alpha =-6.283\ rad/s^2$

Let $\theta$ is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }$

$\theta=\dfrac{-62.83^2}{2\times -6.283}$

$\theta=314.15\ radian$

$\theta=49.99\ revolution$

$\theta=50\ revolution$

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0

3 years ago

In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0

3 years ago

Calcular la energía cinética de un cometa cuya masa es de 5×10 elevado a 31 kg y se mueve con velocidad de 216000km/h

PolarNik [594]

The kinetic energy is $9\cdot 10^{40}J$ .

Explanation:

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

K is the kinetic energy of the object

m is the mass of the object

v is the speed of the object

For the comet in this problem, we have:

$m=5\cdot 10^{31} kg$ is its mass

$v=216,000 km/h$ is the speed

First, we convert the speed from km/h to m/s:

$v=216,000 \frac{km}{h} \cdot \frac{1000 m/km}{3600 s/h}=60,000 m/s$

Therefore, the kinetic energy of the comet is

$K=\frac{1}{2}(5\cdot 10^{31})(60,000)^2=9\cdot 10^{40}J$

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

5 0

3 years ago

A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will

crimeas [40]

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of large apertures the wave passes by or through the obstacle without any significant diffraction.

7 0

3 years ago