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Scilla [17]
3 years ago
8

Suppose a candy bar is 8 cm long, 1 cm high, and 5 cm wide. How many whole candy bars will you be able to fit in a box with a vo

lume of 300 cm3?
A.8 candy bars
B.7.5 candy bars
C.7 candy bars
Physics
1 answer:
Illusion [34]3 years ago
7 0
C: 7 candy bars is the answer i am pretty sure
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A small dog is trained to jump straight up a distance of 1.2m. How much kinetic energy does the 7.2kg dog need to jump this high
Bas_tet [7]
Remember that since energy is conserved, the kinetic energy it takes to jump to a height of 1.2 m is just the same as the potential energy difference between ground level and the 1.2 m height.
Potential energy = (mass)(gravity)(height) = (7.2 kg)(9.81 m/s^2)(1.2 m) = 84.76 J
Therefore, the kinetic energy required for the dog to jump to a 1.2-m height is 84.76 Joules.
5 0
3 years ago
Read 2 more answers
A 5kg rock is lifted 2m. Find the amount of work done. ​
lianna [129]

Answer:

98J

Explanation:

Given parameters:

Mass of rock  = 5kg

Height = 2m

Unknown:

Work done  = ?

Solution:

The amount of work done is given as:

 Work done  = Force x distance

 Work done = Weight x height

  Work done  = mgH

Now insert the parameters and solve;

 Work done  = 5 x 9.8 x 2 = 98J

5 0
2 years ago
Calculate the magnitude and the direction of the resultant forces​
snow_tiger [21]

answer:

resultant = 127.65 in the positive direction

explanation:

F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N

Fy = 50 sin 50 = 50 × -0.26 = -13

Fx = 40 cos 0 = 40×1 = 40

fx = 55 cos 25 = 55×0.99 = 54.45

Fy = 60 sin 70 = 60 × 0.77 = 46.2

resultant = -13+40+54.45+46.2 = 127.65 in the positive direction

6 0
2 years ago
Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t
storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

T=2\pi\sqrt{\frac{I}{K}}

I=mR^{2}

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}

As dimensions of new balance wheel are one-third of their original values

R_{new}=\frac{R}{3}

\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega

5 0
3 years ago
Which refers to an object’s resistance to any change in its motion? force acceleration gravity inertia
satela [25.4K]

<em>Inertia</em> is the property of all matter by which it tends to remain in constant, uniform motion until it's acted on by an external force.

7 0
2 years ago
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