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3 years ago

Why do you think the cup that is exposed to more air has water outside it?

Physics

1 answer:

mylen [45]3 years ago

6 0

It's called condensation. That is what that wetness on the outside of the cup is.

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Suppose that a car starts from rest at t = 0. The car moves with an acceleration of 1.5 m/s2. How far will the car travel in 3.0

rodikova [14]

Answer:

6.75m

Explanation:

To calculate the distance in this question, we can use the formula:

S = ut + 1/2at^2

Where; S = distance

u = initial velocity = 0m/s

t = 3s

a = 1.5m/s^2

Hence:

S = (0 × 3) + 1/2 (1.5 × 3 × 3)

S = 0 + 1/2 (13.5)

S = 13.5/2

S = 6.75

Therefore, the car will travel 6.75m in 3seconds.

3 0

4 years ago

Speed can be represented by many different ____________________ of distance and time.

umka21 [38]

Units is the correct answer

4 0

3 years ago

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

$A=2\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel in terms of the force and extension is,

$Y=\frac{F\times l}{A\times dl}$

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}$

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0

1 year ago

Driving 30.7 m/s in your car you see a dog up ahead and slam on your brakes. You stop just before hitting the dog after skidding

Nitella [24]

Answer: 2.74

Explanation:

We can solve this problem using the stopping distance formula:

$d=\frac{(V_{o})^{2}}{2 \mu g}$

Where:

$d=132.1 m$ is the distance traveled by the car before it stops

$V_{o}=30.7 m/s$ is the car's initial velocity

$\mu$ is the coefficient of friction between the road and the tires

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $\mu$ :

$\mu=\frac{2dg}{(V_{o})^{2}}$

Solving:

$\mu=\frac{2(132.1 m)(9.8 m/s^{2})}{(30.7 m/s)^{2}}$

$\mu=2.74$ This is the coefficient of friction

7 0

3 years ago

What is the answer? quick

elena-s [515]

It would be A. 37.5 because if you multiply 50 and 15 it would be 750 then divide 20 it gives you 37.5.

8 0

3 years ago