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Travka [436]
3 years ago
5

With a force of 5 Newton's Amanda pushes the stacks of books to the right. At the same time Jeremih her little brother pushes th

e stacks of books to the left with a force of 10 Newton's what is the direction and magnitude of the resulting force
Physics
2 answers:
nignag [31]3 years ago
5 0

Answer:

5n to the left

Explanation:

STUISLAND 2020

Fynjy0 [20]3 years ago
4 0

Alright, so solving for the net force is a rather simple and easy step. I want to briefly explain how you determine the net force, but before I do, let's examine the problem together. What has been provided to us? What direction is it going? Well, we can look at the information!

Amanda's Force: 5 N

Jeremiah's Force: 10 N

They're both pushing on the books, but in different directions. Left and right. If Amanda is pushing the books to the right 5 newtons, and Jeremiah is pushing the books to the left with 10 newtons, that means the net force is 5. The book is being pushed to the left! We can say that because Jeremiah's force is much larger than his sisters.

We can determine the net force from the following;

Net Force → 10 N - 5 N = 5 N

In conclusion, your answer should be a total of five (5) newtons

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
3 years ago
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
How much work is done on a satellite in a circular orbit about earth?
alexgriva [62]

Answer:

0 J

Explanation:

W = Work done on the satellite in circular orbit about earth by earth

F = Force on gravity on satellite by earth

d = displacement of the satellite

\theta = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J

3 0
3 years ago
A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free
Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α   .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R      from equation (1)

T = I x a / R²      .... (2)

mg - T = ma    .... (3)

Substitute the value of T from equation (2) in equation (3)

mg - \frac{Ia}{R^{2}}=ma

a=\frac{mg}{m+\frac{I}{R^{2}}}

T is the acceleration in the system

Substitute the value of a in equation (2)

T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}

T=\frac{I\times mg}{I+mR^{2}}

This is the tension in the string.

4 0
3 years ago
A wave oscillates 5.0 times a second and has a speed of 4.0m/s what is the frequency of this wave
Viefleur [7K]

Answer:

The frequency of the wave is 5.0Hz

Explanation:

The frequency of a wave is the number of oscillations or revolutions made in a second.

7 0
3 years ago
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