True or False:

the velocity of a body of mass 60kg reaches 15m/s from 0m/s in 12 second. calculate the kinetic energy and power of the body.

larisa [96]

Answer:Answer

0

Brainly User

Answer:

u = 15m/s

mass =60 kg

v =5 m/s

t = 12 sec

E k= 1/2 ×m v square

= 1/2 (15-5)×(15-5)×60

=1/2 × 10 ×10×60

30 × 100

3000J

power = energy / time

= 3000/ 12

= 250 watt

Explanation:

3 0

3 years ago

What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?

Marta_Voda [28]

Answer:

The answer is $\frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709$ . Let's learn why.

Explanation:

Newton's law of universal gravitation says;

$F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}$

Here G is a universal gravitational <u>constant</u> and is measured experimentally.

Sun's gravitational pull on mercury is:

$F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }$

Therefore $F_{Sun-Mercury} = Gm_{sun} 98,4366$

Sun's gravitational pull on Earth is:

$F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}$

Therefore $F_{Sun-Earth} =Gm_{sun} 265,33$

As a result;

$\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709$

4 0

4 years ago

A yoyo with a mass of m = 150 g is released from rest as shown in the figure.

avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

I is moment of inertia
α is angular acceleration
T is tension in the rope
r is inner radius
R is outer radius
f is frictional force

rT - Rf = Iα ----- (1)

T - f = Ma -------- (2)

a = Rα

where;

a is the linear acceleration of the yoyo

Torque equation for frictional force;

$f = (\frac{r}{R} T) - (\frac{I}{R^2} )a$

solve (1) and (2)

$a = \frac{TR(R - r)}{I + MR^2}$

since the yoyo is pulled in vertical direction, T = mg $a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2$

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0

2 years ago

Sound from a source vibrating at frequency 200 Hz travels at the speed of 300 m/s in air.Wavelength of the sound is (in m/s)

natima [27]

It’s
D)6
Your welcome

6 0

3 years ago

I really need help please just answer at least one

yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

$v_{f}^{2} = v_{i}^{2}-(2*a*x)\\$

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

$v_{f} =v_{i} - (a*t)$

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

$v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})$

Note: The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

$v_{f}= v_{i}+a*t\\$

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

$v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})$

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

$v_{f}= v_{i}+a*t\\$

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0

3 years ago