The convergence of two continental plates would produce

How much force is needed to accelerate at 66 kg skier 4m/s^2 ?

Ray Of Light [21]

E force needed to accelerate a 68 kilogram-skier at a rate of
1.2
m
s
2
is 81.6 Net forces.
Explanation:
Take the mass(68kg) and the acceleration of the skier(
1.2
m
s
2
) and multiply them together

7 0

3 years ago

A 0.91 kg ball and a 2 kg ball are connected by a 0.95 m long rigid, massless rod. The rod is rotating clockwise about its cente

Sloan [31]

Answer: 0.235

Explanation:

Given data.

M1 = 0.91 Kg

M2 = 2kg

L = 0.95m

N = 35rpm

t = 10.44secs

Solution:

• we first solve for the center of gravity

•we calculate our moment of inertia

• we calculate the angular velocity

• we solve for our torque ( t )

6 0

2 years ago

Why dose the moon appear to move frome the east to the west in the night sky?

Gala2k [10]

Answer:

Because Earth rotates on its axis from west to east

Explanation:

5 0

3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

An open train car moves with speed 18.5 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to

olga2289 [7]

Answer:

The speed decreases.

Explanation:

This can be explained using the conservation of linear momentum.

Since there is no friction, the initial moment of the train must be equal to its linear moment after it is filled with water.

the initial linear momentum is

$m_{1}v_{1}$

where $m_{1}$ is the initial mass of the train, and $v_{1}$ the initial speed of the train.

And linear momentum after the water filled the train car is

$m_{2}v_{2}$

where $m_{2}$ is mass of the train after the rain, and $v_{2}$ the speed of the train after the rain

<u>the equality must be fulfilled:</u>

$m_{1}v_{1}=m_{2}v_{2}$

We know that if water is added to the train, $m_{2}$ that is the mass after the water is added, is greater than $m_{1}$ which is the mass of the train without the water.

Therefore, in order for the conservation of the linear momentum to be fulfilled: $m_{1}v_{1}=m_{2}v_{2}$

the speed after the water is added ( $v_{2}$ ) must be smaller than the initial train speed ( $v_{1}$ ) . So the speed of the car decreases.

3 0

3 years ago