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3 years ago

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1. A person is lifting a suitcase into the trunk of a car. Assuming the suitcase is already in the air, draw the FBD and write t

he sum of forces equation for the box. (Assume the person is lifting straight up).

1 answer:

Alik [6]3 years ago

7 0

Answer:

Explanation:Assume you are lifting an object with mass 20 kg from the ground to a height of 1.5 m. Assume that you are exerting a constant force in the upward direction and that you are moving the object upward with uniform velocity. The net force on the object is zero. The force you are exerting is equal in magnitude and opposite in direction to the force of gravity. As you are lifting the object you are doing work on the object.

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An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

3 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago

Please help me!!!!!!!! it's due soon!!!!!!!!!!!!! NO LINK!!!!!!!!!

maria [59]

.75(m/s)^2

Use this picture
If you know the two on the bottom you multiply them but if you only know the top and one on the bottom you divide

7 0

3 years ago

What is the velocity in meters per second of a runner who runs exactly 110 m toward the beach in 72 seconds.

KATRIN_1 [288]

1.53 m/s toward the beach

Explanation:

The magnitude of the velocity of the runner is given by:

$v=\frac{d}{t}$

where

d is the displacement of the runner

t is the time taken

In this case, d=110 m and t=72 s, so the velocity of the runner is

$v=\frac{110 m}{72 s}=1.53 m/s$

Velocity is a vector, so it consists of both magnitude and direction: we already calculate the magnitude, while the direction is given by the problem, toward the beach.

3 0

3 years ago

Read 2 more answers

If you exert a force on an object in motion you will change its ____.

devlian [24]

If you exert a force on an object in motion, then depending on the
direction of the force you exert and the direction in which it's already
moving, you may speed it up, slow it down, or change the direction
of its motion. Any of these changes is called an acceleration.

In addition to that, you'll change the object's momentum and kinetic energy.
They may increase or decrease ... again depending on the directions of the
motion and the new force.

You will not change the object's mass, inertia, weight, color, cost,
political affiliation, or gender preference.

4 0

3 years ago

Read 2 more answers