A column of water exerts a pressure of 100.Pa and a force of 2.40 N. What is the cross-sectional area of a the column?

An aeroplane accelerates from rest to 65 m/s for take-off. It travels for

Virty [35]

Answer:

1.76m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity = 65m/s

Distance traveled = 1200m

Unknown:

Acceleration = ?

Solution:

This is linear velocity and we apply the appropriate motion equation to solve this problem.

V² = U² + 2as

S is the distance

u is the initial velocity

V is the final velocity

a is the acceleration

Now, insert the parameters and solve;

65² = 0² + 2 x a x 1200

4225 = 2400a

a = 1.76m/s²

5 0

3 years ago

How much gravitational force do two lead balls with a mass of 8 kilograms, the centers of mass of which are 17 cm apart, affect

Arisa [49]

Answer:

1.48×10⁻⁷ Newtons

Explanation:

From the question,

According to newton's law of universal gravitation.

F = Gmm'/r²........................ Equation 1

F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.

Given: m = m' = 8 kg, r = 17 cm = 0.17 m,

Constant : G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F = (6.67×10⁻¹¹×8×8)/(0.17²)

F = 1.48×10⁻⁷ N

7 0

3 years ago

Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7

Dmitry [639]

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

$v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}$ ......(I)

The orbital speed of satellite B,

$v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}$ ......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}$

Put the value into the formula

$\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}$

$\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}$

Hence, The ratio of their orbital speeds are 5:4.

8 0

3 years ago

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

5 0

3 years ago

Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is

Naddika [18.5K]

Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

$a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2$

Therefore, the rate of acceleration of the train is 4 m/s²

5 0

3 years ago