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Bad White [126]
3 years ago
14

What does the area between the line and the x-axis represent on a velocity vs. time graph?

Physics
2 answers:
Finger [1]3 years ago
7 0

Answer:

It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.

Explanation:

Rashid [163]3 years ago
5 0

Answer: It represents displacement

Explanation:

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A spring is used as part of a lift system and follows Hooke's law. If the spring is
salantis [7]

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

6 0
3 years ago
The section of the electromagnetic spectrum that humans can generally see is called _____ light.
Lisa [10]

Answer:

(A) Visible

Explanation:

  • The section of the electromagnetic spectrum that humans can generally see is called visible light.

White light is visible light and the range of visible wavelengths ranges from 400 - 700 nanometers.

8 0
4 years ago
Read 2 more answers
The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec
Elanso [62]
On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
where L is the length of the pendulum and g_{earth}=9.81~m/s^2 is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
6 0
3 years ago
The correct answer iS: 3.60 s
Rainbow [258]

Answer: Your answer is 7.77 m/s

Explanation:

Yw and mark me as brainiest

3 0
2 years ago
a rectangular coil of 25 loops is suspended in a field of 0.20wb/m2.the plane of coil is parallel to the direction of the field
statuscvo [17]

Answer:

The current in the coil is 60 Ampere.

Explanation:

Given:

Number of turns in the coil is N = 25

Dimension of the coil = 15cm X 12cm

magnitude of magnetic field = 0.20T

angle in the xy plane is θ = 0 degree

torque τ = 5.4 N-m

To find:

current in the coil is i = ?

Solution:

The torque acting on the coil is given by

=> \tau = NiAB cos\theta

Converting cm to m

12 cm = 0.12 m

15 cm = 0.15 m

The area of the coil is  

A = 0.12 X 0.15

A = 0.018 m^2

Substituting the values

=>5.4 = 25\times i \times 0.018 \times 0.20 \times cos\theta

=>5.4 = 25\times i \times 0.018 \times 0.20 \times cos(0)

=>5.4 = 25\times i \times 0.018 \times 0.20 \times 1

=>5.4 = 25\times i \times 0.018 \times 0.20 \times 1

=>5.4 = 0.09\times i

=>i = \frac{5.4}{0.09}

=> i = 60 A

3 0
4 years ago
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