Which type of rock can most easily be split into thin sheets?

Objects moving along a circular path have a centripetal accel- eration provided by a net force directed towards the center. Iden

Andrews [41]

Answer:

a)

The distance between planet and the sun is more that is only gravitational force will act between them.

The gravitational force which act between two planet.

b)

When a car going around a friction force act on the car towards the center.

c)

When the rock at the highest position the tension and gravitational force will act.

d)

The wall of the dryer will apply the normal force.

5 0

4 years ago

Read 2 more answers

A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta

-BARSIC- [3]

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

= 107 cos 13.5°= 104 N

Power = $\dfrac{work\ done}{time}$

69.4 = $\dfrac{W}{33}$

W = 2290.2 J

Work done= Force x displacement

displacement= $\dfrac{2290.2}{104}$

= 22.02 m

6 0

3 years ago

In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc

Harman [31]

Answer:

$A=\frac{A_0}{\sqrt 3}$

Explanation:

Initial amplitude= $A_0$

We are given that

Amplitude after 2 s=A= $\frac{1}{3}A_0$

We have to find the amplitude after 1 s.

We know that amplitude at any time t

$A=A_0e^{-\alpha t}$

Using the formula

$\frac{A_0}{3}=A_0e^{-2\alpha}$

$\frac{1}{3}=e^{-2\alpha}$

$3=e^{2\alpha}$

$ln 3=2\alpha$

$\alpha =\frac{ln 3}{2}=ln\sqrt 3$

$e^{\alpha}=\sqrt 3}$

When t=1 s

$A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}$

8 0

4 years ago

HELP ASAP am being timed will brain crown mark!

Nana76 [90]

Answer:

The answer is D.

Explanation:

This is because mass always remains constant and weight is dependent on a gravitational pull.

3 0

4 years ago

Un coche de carreras de Fórmula 1 (véase Figura 2.30) acelera desde el reposo a razón de 18 m.s-2 . Suma que se mueve en línea r

Lynna [10]

Answer:

I will answer in English.

Ok, we know that the acceleration is a = 18m/s^2, and we have that the initial velocity and position are both zero. (because it starts at rest)

then we have:

a(t) = 18m/s^2

for the velocity, we integrate over time (because the initial velocity is equal to zero we do not have any integration constant)

v(t) = (18m/s^2)*t

for the position we integrate again over time, and again, we do not have any integration constant

p(t) = (1/2)(18m/s^2)*t^2 = (9m/s^2)*t^2

a) The speed at t= 3s can be found by replacing t = 3s in the velocity equation.

v(3s) = (18m/s^2)*3s = 54m/s

b) the distance traveled by this time can be found by replacing t = 3s in the position equation.

p(3s) =(9m/s^2)*(3s)^2 = 81 m

c) first, we need to find what is the time when the position is equal to 200m.

p(t) = 200m = (9m/s^2)*t^2

√(200/9) s = t = 4.7s

Now we replace that time in the velocity equation and we get:

v (4.7s) = (18m/s^2)*4.7s = 84.6m/s

d) ok, to do this we know that.

1 hour has: 60*60 = 3600 seconds.

then we have the transformation k = 1h/3600s

1 km has 1000 meters.

then we have the transformation c = 1km/1000m

so we have that:

84.6m/s = 84.6m/s*(c/k) = 84.6*(3600/1000)km/h = 304.56 km/h

6 0

3 years ago