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Veseljchak [2.6K]
3 years ago
14

Anyone can help?? I need it done before 9am please!!

Physics
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

The answer is below

Explanation:

The equation for a linear line graph is given by:

y = mx + b, where y and x are variables, m is the slope of the graph and b is the y intercept (that is value of y when x is zero).

The slope (m) of a line passing through two points (x_1,y_1)\ and\ (x_2,y_2) is given by:

m=\frac{y_2-y_1}{x_2-x_1}

A) The first line passes through the point (0, 0) and (10, 60). It is represented as (time, velocity). Hence the slope is:

m=\frac{60 -0}{10-0}=6\ m/min^2

B) The second line passes through the point (10, 60) and (15, 60). Hence the slope is:

m=\frac{60 -60}{15-10}=0\ m/min^2

C) The third line passes through the point (15, 60) and (40, -40). Hence the slope is:

m=\frac{-40 -60}{40-15}=-4\ m/min^2

D) The third line passes through the point (40, -40) and (55, 0). Hence the slope is:

m=\frac{0 -(-40)}{55-40}=2.67\ m/min^2

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7 0
3 years ago
A ball, initially at rest, reaches a speed of 20 m/s at the bottom of the apparatus. If it takes the ball 10 seconds to reach th
MariettaO [177]

Answer:

2 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 20 m/s

t = 10 s

Find: a

a = (v − v₀) / t

a = (20 m/s − 0 m/s) / 10 s

a = 2 m/s²

3 0
3 years ago
PLEASE HELP (100)
morpeh [17]

Answer:

The work done will be W=55.27\: J

Explanation:

The work equation is given by:

W=F\cdot x

Where:

F is the force due to gravity (weight = mg)

x is the length of the ramp (3 m)

Now, the force acting here is the component of weight in the ramp direction, so it will be:

F_{x-direction}=mgsin(28)

Therefore, the work done will be:

W=mgsin(28)*3

W=4*9.81*sin(28)*3

W=55.27\: J

I hope it helps you!

3 0
3 years ago
Which of these would represent the force of gravity
tigry1 [53]

Missing figure: find it in attachment.

Answer:

Force D

Explanation:

In order to answer the question, let's keep in mind that the force of gravity on an object on Earth is the attractive force exerted by the Earth on the object; its direction is always downward (towards the Earth's centre), and its magnitude is given by

F = mg

where m is the mass of the object and g is the acceleration of gravity.

It follows immediately that in the figure, the force of gravity is the only force acting downward: therefore, force D.

The other forces are called:

Force A: thrust (it is the forward force generated by the engines)

Force B: lift (it is the upward produced by the aerodynamics of the wings)

Force C: air resistance (it is the backward force due to the friction between the air and the surface of the plane)

7 0
3 years ago
A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of
Aleksandr [31]

Answer:

Part 1)

L_1 = 185.2 kg m^2/s^2

Part 2)

L_2 = 663.07 kg m^2/s^2

Part 3)

L = 663.07 kg m^2/s^2

Part 4)

\omega = 1.83 rad/s

Part 5)

F_c = 453.6 N

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

L_1 = I \omega

here we know that

I = 194 kg m^2

\omega = 1.47 rad/s^2

now we have

L_1 = 194 \times 1.47

L_1 = 185.2 kg m^2/s^2

Part b)

Angular momentum of the person is given as

L = mvR

so we have

m = 68 kg

v = 4.9 m/s

R = 1.99 m

so we have

L_2 = (68)(4.9)(1.99)

L_2 = 663.07 kg m^2/s^2

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

L = 663.07 kg m^2/s^2

Part 4)

By angular momentum conservation of the system we will have

L_1 + L_2 = (I_1 + I_2)\omega

185.2 + 663.07 = (194 + 68(1.99^2))\omega

848.27 = 463.28 \omega

\omega = 1.83 rad/s

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

F_c = m\omega^2 R

F_c = 68(1.83^2)(1.99)

F_c = 453.6 N

3 0
3 years ago
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