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3 years ago

Three dogs are fighting over a bone. One is pulling to the left with a force of 20N, another

Physics

1 answer:

Len [333]3 years ago

8 0

Answer:

Net force = 21.12 N

angle for the force α = 45°

Explanation:

In this problem, we will take the left as the negative x-axis, the right as the positive x-axis, and the upward direction as the positive y-axis.

The 20 N force to the left has an x component of -20 N, and y component of 0 N

The 35 N force to the right has a x component of 35 N, and a y component of 0 N

The 15 N upwards has an x component of 0 N, and a y component of 15 N

We resolve the forces into the x and y components.

for the x component Fx

Fx = -20 + 35 + 0 = 15 N

For the y component Fy

Fy = 0 + 0 + 15 = 15 N

Net force Fn = $\sqrt{Fx^{2} + Fy^{2} }$

$Fn = \sqrt{(15)^{2} + (15^{2} ) }$

Net force Fn = 21.21 N

for the angle,

$tan\alpha = \frac{Fy}{Fx} = \frac{15}{15}$

$tan\alpha = 1$

α = $tan^{-1}$ 1

angle for force α = 45°

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The table shows information about four students who are running around a track. Which statement is supported by the information

murzikaleks [220]

Answer:

Mohammed has less kinetic energy than Autumn

Explanation:

The kinetic energy of each student is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the student

v is the speed of the student

Let's use the formula above to calculate the kinetic energy of each student:

- Autumn: $K=\frac{1}{2}(50 kg)(4 m/s)^2=400 J$

- Mohammed: $K=\frac{1}{2}(57 kg)(3 m/s)^2=256.5 J$

- Lexy: $K=\frac{1}{2}(53 kg)(3 m/s)^2=238.5 J$

- Chiang: $K=\frac{1}{2}(64 kg)(5 m/s)^2=800 J$

Therefore, by looking at the numbers, we see that the correct answer is

Mohammed has less kinetic energy than Autumn

7 0

3 years ago

Read 2 more answers

You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?

Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 = v0 + g · t

t = -v0/g

Now, replacing t = -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g · (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g). Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

7 0

4 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$