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Alisiya [41]
2 years ago
15

11. Velocity-time graph for the motion of an object in a straight path is a straight line parallel to the time axis

Physics
1 answer:
Vilka [71]2 years ago
8 0

a) uniform velocity

b) zero or no acceleration

c) (see picture)

EXPLANATION:

(see picture)

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A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 65.5-gram mass is attached at the 21.0-cm
Gnesinka [82]

Answer:

113.53 g

Explanation:

Please see attached photo for explanation.

In the attached photo, M is the mass of the meter stick.

The value of M can be obtained as shown below:

Clockwise moment = M × 10.5

Anticlockwise moment = 65.5 × 18.2

Anticlockwise moment = Clockwise moment

65.5 × 18.2 = M × 10.5

1192.1 = M × 10.5

Divide both side by 10.5

M = 1192.1 / 10.5

M = 113.53 g

Thus, the mass of the meter stick is 113.53 g

7 0
3 years ago
Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small
ahrayia [7]
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by 

<span>C = 2L = 2*pi*R ---> R = L/pi </span>

<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>

<span>we can define a small charge dq as </span>

<span>dq = l*ds = l*R*da </span>

<span>So the electric field can be written as: </span>

<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>

<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>

<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>

<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
8 0
3 years ago
If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules,
Kitty [74]

Answer:

352,088.37888Joules

Explanation:

Complete question;

A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.

A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)

Chane in potential energy is expressed as;

ΔGPH = mgΔH

m is the mass of the hiker

g is the acceleration due to gravity;

ΔH is the change in height

Given

m = 53kg

g = 9.8m/s²

ΔH = 2574-350 = 2224ft

since 1ft = 0.3048m

2224ft = (2224*0.3048)m = 677.8752m

Required

Gravitational potential energy

Substitute the values into the formula;

ΔGPH = mgΔH

ΔGPH = 53(9.8)(677.8752)

ΔGPH = 352,088.37888Joules

Hence the gravitational potential energy is 352,088.37888Joules

7 0
3 years ago
What is the name for the bunched up region of<br> air particles in the sound wave?
Leviafan [203]

Answer:

These regions are known as compressions and rarefactions respectively. The compressions are regions of high air pressure while the rarefactions are regions of low air pressure.

Explanation:

4 0
3 years ago
Read 2 more answers
A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0
3 years ago
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