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3 years ago

8

Several motorboats with the same mass are used in an experiment. The forces of the different motors versus their accelerations a

re graphed. What is the y-intercept of this graph? (1point) Does anyone know the answer?

1 answer:

Yuliya22 [10]3 years ago

7 0

Answer:

C. the velocity

Explanation:

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How many significant figures does the following number have: 0.002040?

ANTONII [103]

Answer: 4

Explanation: because 0s aren’t significant and after the decimal point, there was to be a value greater than 0 than the rest are sig figs.

5 0

3 years ago

A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

4 0

4 years ago

An object has a kinetic energy of 275 j and a momentum of magnitude 25.0 kg · m/s. find the (a) speed and (b) mass of the object

Sergeu [11.5K]

Kinetic energy = momentum^2 / 2 x mass
Mass = (momentum^2/ Kinetic energy) / 2

Mass = (25^2 / 275) / 2
Mass = 1.136 kg.

momentum = mass x velocity

velocity = mass / momentum
velocity = 1.136 / 25
velocity = 0.04544 m/s

8 0

3 years ago

Read 2 more answers

Petroleum is a nonrenewable resource. true or false

Nata [24]

Coal, petroleum, and natural gas are considered nonrenewable because they can not be replenished in a short period of time

5 0

3 years ago

Based on Newton's law of motion, which combination of rocket bodies and engine will result in the acceleration of 40 m/s ^2 at t

ad-work [718]

The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.

Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?

Body 3 + Engine 1

Body 2 + Engine 2

Body 1 + Engine 2

Body 1 + Engine 1

Solution :

Given :

Body Mass (gram) Engine Force (newtons)

1 500 1 25

2 1500 2 20

3 750 3 30

The body 1 has a mass of 500 gram which is equal to 0.5 kg

And engine 2 has a force of 20 newtons.

We know that according to Newton's laws of motion,

Force = mass x acceleration

20 = 0.5 x acceleration

Acceleration $=\frac{20}{0.5}$

$=\frac{200}{5}$

$= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $40 \ m/s^2$ at the start of the launch.

8 0

3 years ago