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faltersainse [42]
3 years ago
8

A rectangular object was found to have a mass of 1.278 kg and density of 4.98  g/cm3. Suppose that you knew that the length was

47 mm and the width was 61 mm. Using this information, compute the height of the rectangle in cm.
Physics
1 answer:
Leni [432]3 years ago
5 0

Answer:

89.6 cm

Explanation:

From the question,

Volume of the rectangular object = Mass/Density.

V = m/D.................. Equation 1

Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³

Substitute into equation 1

V = 1.278/4980

V = 2.57×10⁻⁴ m³.

But,

V = lwh............... Equation 2

Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.

make h the subject of the equation

h = V/lw........... Equation 3

Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.

Substitute into equation 3

h = 2.57×10⁻⁴/(0.047×0.061)

h = 0.896 m

h = 89.6 cm

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2 years ago
Define pressure in short
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Answer:

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Explanation:

Pressure, in the physical sciences, the perpendicular force per unit area, or the stress at a point within a confined fluid.

7 0
3 years ago
What is the amplitude of this wave?
nikdorinn [45]

Answer:

2.1 cm

Explanation:

The general equation for the graph is:

y = A cos(ωt + ∅)

A: amplitude, is the maximum and minimum value for y(t)

5 0
3 years ago
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
List the main types of electromagnetic waves in order of increasing frequency. radio waves, microwaves, infrared, visible light,
alexandr1967 [171]

Answer:

radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays

Explanation:

Electromagnetic waves are oscillations of electric and magnetic fields in a direction perpendicular to the direction of motion of the wave (transverse waves). They are classified into 7 different types, according to their frequencies.

From lowest to highest frequency, we have:

Radio waves

Microwaves 10^9 Hz - 4\cdot 10^{13}Hz

Infrared 4\cdot 10^{13} - 4\cdot 10^{14} Hz

Visible light 4\cdot 10^{14} - 8\cdot 10^{14}Hz

Ultraviolet 8\cdot 10^{14} - 2.4\cdot 10^{16} Hz

X-rays 2.4\cdot 10^{16} -5 \cdot 10^{19}Hz

Gamma rays >5\cdot 10^{19} Hz

3 0
4 years ago
Read 2 more answers
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