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3 years ago

A 43 kg box is being pushed and pulled by

Physics

1 answer:

m_a_m_a [10]3 years ago

4 0

Answer:

a = - 0.209 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

We will take the positive forces to the right and the negative forces to the left.

$155-277+113=43*a\\-9=43*a\\a = - 0.209[m/s^{2} ]$

The negative sign means that the box accelerates in a negative direction to the left.

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A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3 km/h.

Amanda [17]

Answer:

$F_{1}=\frac{1}{5}F_{2}$ or $F_{2}=5F_{1}$

In other words, $F_{1}$ is one fifth of $F_{2}$ or $F_{2}$ is five times as big as $F_{1}$

Explanation:

In order to solve this problem we must start by sketching the situation (refer to the attached picture).

When the ship is pulled only by force 1, it will change its speed by 3km/hr in 10 seconds. So in order to use these values we need to either turn the km/hr in km/s or turn the seconds to hours. Let's turn the seconds to hours:

$10s*\frac{1hr}{3600s}=\frac{1}{360} hr$

so we can now use the acceleration formula to find the acceleration of the boat so we get:

$a=\frac{\Delta v}{\Delta t}$

which will give us an accceleration of:

$a=\frac{3km/hr}{\frac{1}{360}hr}=1080km/hr^{2}$

once we got the acceleration we can for sure say taht:

$F_{1}=ma=m*1080\frac{km}{hr^{2}}$

Now, if we take a look at the second drawing we can see that the resultant force applied to the boat is found by adding the two forces, force one and force two, so we get:

$F_{1}+F_{2}=ma$

in this case the acceleration changes because the change in velocity is of 18km/hr in the same 10 seconds, so we get that:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{18km/hr}{\frac{1}{360}hr}=6480km/hr^{2}$

so we can say that:

$F_{1}+F_{2}=m*6480km/hr^{2}$

we can substitute the first force into this equation so we get:

$m*1080km/hr^{2}+F_{2}=m*6480km/hr^{2}$

and solve for the second force, so we get:

$F_{2}=m*6480km/hr^{2}-m*1080km/hr^{2}$

which yields:

$F_{2}=m*5400km/hr^{2}$

Now we can compare theh two forces, force 1 and force 2 by dividing them:

$\frac{F_{1}}{F_{2}}=\frac{m*1080km/hr^{2}}{m*5400km/hr^{2}}$

which yields:

$\frac{F_{1}}{F_{2}}=\frac{1}{5}$

when solving for the first force we get:

$F_{1}=\frac{1}{5}F_{2}$

which tells us that the second force is one fifth of the first force.

and when solving for the second force we get that:

$F_{2}=5F_{1}$

which means that the second force is 5 times as big as the first force.

8 0

3 years ago

Please help me i neeed help with all threeas soon as possible thank you

liberstina [14]

Answer:

4b: comets

5a: supercluster

5b: they just changed 4b's solar system for milky way. I think it is still comets. if not, then just say black holes.

7 0

3 years ago

A 12000 kg train engine moving at 2.2 m/s hits and locks into 3 boxcars with a total mass of 25000 kg sitting still. If the coll

PilotLPTM [1.2K]

Answer:

V = 0.714m/s

Explanation:

Full solution calculation can be found in the attachment below.

From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.

Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.

See the attachment below for the solution calculation.

5 0

3 years ago

Read 2 more answers

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposi

trasher [3.6K]

Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

3 0

3 years ago

HELP PLEASE!

Alborosie

What is this on, is this on a test?

4 0

3 years ago

Read 2 more answers