An example of a scientific measurement when determining the period of a pendulum is

How can I connect projectile motion into inclines. plsss give me ideas

Alenkasestr [34]

Answer:

Explanation:

Initial launch angle, θ

Initial velocity, u.

Time of flight, T.

Acceleration, a.

Horizontal velocity, vx.

Vertical velocity, vy.

Displacement, d.

Maximum height, H.

4 0

4 years ago

How is the potential difference same in capacitors arranged in parallel combination?

zimovet [89]

Answer:

<h2>Potential difference across capacitors in parallel </h2>

Two or more capacitors are said to be connected in parallel if each one of them is connected across the same two points. In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor will store different charge.

8 0

3 years ago

Pls help me pls<br>pls use the points as my thank you

Ivanshal [37]

Answer:

A

Explanation:

In section A, she is moving in constant speed because there's a flat line which indicates speed remained same as time passed. In section c, there's a flat line but look at y axis for this one, it is 0 which means there's no speed during this part. B and D have varying speeds as there is a change in y axis variables as time passes.

4 0

3 years ago

Read 2 more answers

A 35 kg trunk is pushed 4.8 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic fric

netineya [11]

Answer:

We need to separate the x- and y-components of the applied force. For simplicity, I will denote the direction along the inclined plane as x-direction, and the perpendicular direction as y-direction.

$F_x = F\cos(30^\circ)\\F_y = F\sin(30^\circ)$

Only the x-component of the applied horizontal force does work on the trunk.

But we need to find the magnitude of the force. We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

$0 = F\cos(30^\circ) - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0.86F = 35(9.8)(0.5) + 35(9.8)(0.86)(0.19)\\F =264.5N$

$F_x = 264.5\cos(30^\circ) = 227.5N\\F_y = 264.5\sin(30^\circ) = 132.25N$

$W_{F_x} = F_xd = 227.5\times 4.8 = 1092J$

The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

$W_g = -mg\sin(30^\circ)d$

Note that the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

$W_g = -823J$

The energy dissipated by the frictional force can be found as follows:

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Additionally, the sum of work done by the friction and weight is equal in magnitude to the work done by the applied force. This shows that our calculations are consistent.

In the second part of the question, the applied force is on the x-direction. We will follow a similar procedure but a different force.

$0 = F - mg\sin(30^\circ) - mg\cos(30^\circ)\mu\\0 = F - 35(9.8)(0.5) - 35(9.8)(0.86)(0.19)\\0 = F - 171.5 - 56\\F = 227.5N$

$W_F = Fd = 227.5\times 4.8 = 1092J$

$W_g = -mg\sin(30^\circ)d = -35(9.8)(0.5)(4.8) = -823J$

$W_f = -mg\cos(30^\circ)\mu d = -35(9.8)(0.86)(0.19)(4.8) = -269J$

Explanation:

As you can see above, the answers are the same, although the directions of the applied forces are different. The reason for this situation is that in the first part the y-component does no work.

8 0

4 years ago

True or false: The force that causes a satellite to orbit Earth is gravity.

Kisachek [45]

The force that causes a satellite to orbit Earth is gravity ------ True

4 0

4 years ago