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3 years ago

When a car rolls down a ramp, which force supports the car's weight?

Physics

1 answer:

jasenka [17]3 years ago

6 0

The force of gravity most likely

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What happens when the temperature of liquid water drops

ser-zykov [4K]

Answer: turns to gas

Explanation:

when a liquid gets heated up the chemicals start to boil/evaporate turning the liquid into a gas

5 0

3 years ago

I need help on this science

Serga [27]

Ok ok ok ok ok ok ok ok ok

8 0

3 years ago

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

4 0

3 years ago

A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu

Murljashka [212]

The work done on the ship is $1.5\cdot 10^7 J$

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we have:

$F=5.00\cdot 10^3 N$ (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

$\theta=0^{\circ}$ (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

$W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0

3 years ago

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago