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2 years ago

7

If a body is moving in the horizontal axis with a velocity Vx= 6m/s and in the vertical axis Vy=8m/s What is the angle Theta abo

ve the horizontal axis of the velocity of the body V? A. Theta = 45.0 degrees B.Theta = 63.3 degrees C. Theta = 53.1 degrees D. Theta = 43.7 degrees

1 answer:

cluponka [151]2 years ago

7 0

Answer: C

Explanation: It's a lot of math.

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Rod A has twice the diameter of rod B, but both are made of iron and have the same initial length. Both rods are now subjected t

Vilka [71]

Answer:

ΔLa/ΔLb = 1

Explanation:

The change in length of a solid is given by the following formula:

ΔL = α L ΔT

where,

ΔL = Change in length

α = coefficient of linear expansion

L = Original Length

ΔT = Change in Temperature

Since, the length and change in temperature for both rods are same. Also, the material of each rod is same, which implies that coefficient of linear expansion for both rods is same. Hence, the ratio of change in length of both rods will be:

<u>ΔLa/ΔLb = 1</u>

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2 years ago

In ocean waves, water particles move ________ and energy moves ________.

morpeh [17]

In ocean waves, water particles move with mechanical energy and energy moves with gravity

Not sure but hope it helps!

6 0

1 year ago

Read 2 more answers

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sl

KonstantinChe [14]

Answer:

Acceleration, $a=1.2\ m/s^2$

Explanation:

Given that,

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{240\ N}{2\times 100\ kg}$ (m = 2m)

$a=1.2\ m/s^2$

So, the acceleration of the sleds is $1.2\ m/s^2$ .

6 0

2 years ago

3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg

Anni [7]

Answer:

$\mathbf{1.51\times10^{-15}N}$

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

$r=1cm=0.01m$

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

$\Rightarrow W=mg$

$=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}$

$=\mathbf{1.51\times10^{-15}N}$

6 0

1 year ago

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

enyata [817]

Answer:

Emitted power will be equal to $7.85\times 10^{-5}watt$

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing $I=10^{-12}W/m^2$

We have to find the emitted power

Emitted power is equal to $P=I\times A$

$P=I\times 4\pi R^2$

$P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt$

So emitted power will be equal to $7.85\times 10^{-5}watt$

4 0

1 year ago