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masha68 [24]
3 years ago
7

Please help me-. it’s part of my major grade.

Physics
2 answers:
stich3 [128]3 years ago
7 0
It will continue to not move- if you leave an object alone it’s not going to move unless something happens to it - I.e an outside force.

Let me know if you have any questions
Serga [27]3 years ago
6 0

Answer:

c

Explanation:

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Someone pls help me will give brainlist
Crazy boy [7]

Answer:

The volume is 5.6 L

The mass is 100.2g

Explanation:

Quantitative observations are more based on numbers and values

6 0
3 years ago
An object initially at rest experiences an acceleration of 1.90 ­m/s² for 6.60 s then travels at that constant velocity for anot
borishaifa [10]

Answer:

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

Explanation:

Hi there!

The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.

The displacement is calculated as the distance between the final position of the object and the initial position. <u>In this problem</u>, the displacement is equal to the traveled distance because the object travels only in one direction:

a.v = Δx/t

Where:

a.v = average velocity.

Δx = displacement = final position - initial position

t = time

So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:

x = 1/2 · a · t²

x = 1/2 · 1.90 m/s² · (6.60 s)²

x = 41.4 m

The object traveled 41.4 m during the first 6.60 s.

Now, let's find the rest of the traveled distance.

When the velocity is constant, a = 0. Then, the equation of position will be:

x = x0 + v · t

Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:

x = v · t

The velocity reached by the object during the acceleration phase is calculated as follows:

v = v0 + a · t   (v0 = 0 because the object started from rest)

v = 1.90 m/s² · 6.60 s

v = 12.5 m/s

Then, the distance traveled by the object at a constant velocity will be:

x = 12.5 m/s · 8.50 s

x = 106 m

The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.

Then the displacement will be:

Δx = final position - initial position

Δx = 147 m - 0 = 147 m

and the average velocity will be:

a.v = Δx/t

a.v = 147 m / 15.1 s

a.v = 9.74 m/s

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

8 0
3 years ago
An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free
Blababa [14]

Answer:

the free-fall acceleration on the moon is 1.68 m/s^2

Explanation:

recall the formula for the gravitational potential energy (under acceleration of gravity "g"):

PE = m * g * h

replacing with our values for the problem:

46 J = 91 * g * 0.3

solve for the "g" on the Moon:

g = 46 / (91 * 0.3)

g = 1.68  m/s^2

3 0
3 years ago
A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the
levacccp [35]

Tension in the rope due to applied force will be given as

F = 142 N

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

W = F d cos\theta

now we can plug in all values above

W = 142 * 6.1 * cos37

W = 691.8 J

So here work done to pull is given by 691.8 J


8 0
3 years ago
Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.
Nikolay [14]
I think answer should be d. Please give me brainlest let me know if it’s correct
5 0
3 years ago
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