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Dvinal [7]
2 years ago
14

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car

Physics
1 answer:
Novay_Z [31]2 years ago
0 0

Answer:

4.5 *10^{3}\ kgm/s^{2}

Explanation:

Given:

Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\  meters/second^{2}

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration  we get ,

Force\  = \ 1.00 * 10^{3} * 4.5\ m/s^{2}

Force\  =4.5 * \ 10^{3} \  kgm/s^{2}

Therefore Net force is :4.5 *10^{3}\ kgm/s^{2}

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Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad
mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = \rhoV

V = Volume =\dfrac{4}{3}\pi r^3

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

\rho = Density = 19 g/cm³

c = Speed of light = 3\times 10^8\ m/s

A = Area = \pi R^2

I = Intensity = \dfrac{P}{4\pi r^2}

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}

According to the question

F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W

The power required of the light source is 468449163762.0812 W

4 0
2 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
2 years ago
Describe several uses of plastic, and explain why plastic is a good choice for these products
miv72 [106K]
Plastic is used in many things like bags, toys, nd certain products. Plastic is very durable nd flexible :) hope i helped a bit c;
5 0
2 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati
fgiga [73]

Answer:

5.5 rad/sec

Explanation:

8 0
2 years ago
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