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mixas84 [53]
3 years ago
12

an object starts from rest and travels 100 m in 10 s. what will be the velocity of the object after 10 s

Physics
1 answer:
astra-53 [7]3 years ago
6 0

Answer:

10m/s (if the velocity is constant!!)

Explanation:

v=Δx/Δt=100/10=10m/s (if the velocity is constant!!) (i dont know if the letters are right in english sorry)

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What is another unit for momentum besides kg-m/s?<br> a. N<br> b. N-s<br> c. N-s2<br> d. N/s
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'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.
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3 years ago
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
pochemuha

Answer:

m = 63 grams

Explanation:

ω = 10 cycles/s(2π radians/cycle) = 20π rad/s

ω = √(k/m)

m = k/ω² = 250/(20π)² = 0.06332... kg

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3 years ago
What is the term for the process by which a portion of a glacier breaks off and falls into the water
zhenek [66]

The term for the process by which a portion of a glacier breaks off and falls into the water is called calving.

6 0
3 years ago
A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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3 years ago
An object moves from point A to point C along the rectangle shown in the figure below.
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Answer:

Hello friend where is the figure of the question

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