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2 years ago

ILL MARK BRAINLIEST PLEASE HELP 2.Write the formula associating energy with power

Physics

1 answer:

bekas [8.4K]2 years ago

7 0

Answer:

The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.

Explanation:

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A car driving at a constant speed of 64 mi/h travels 68 miles. How many hours did this take?

babymother [125]

64 miles/hour
Therefore 1/64 hours/mile
68 miles * 1/64 hours/mile (notice how miles cancels out)
Therefore the answer is 68/64 hours = 1.0625 hours = 1 hour 3min and 45sec.

3 0

3 years ago

A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : $f=20\ cm$

Height of object : $h=1\ cm$

Object distabce from lens : $u=-10\ cm$

Using lens formula: $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ , we get

$\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}$ , where v = image distance from the lens.

On solving aboive equation , we get

$\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm$

Formula of Magnification : $m=\dfrac{v}{u}=\dfrac{h'}{h}$ , where h' is the height of image.

Put value of u, v and h in it , we get

$\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm$

Hence, the height of the image is 2 cm.

3 0

2 years ago

How does freezing and evaporation affect salinity?

Gnoma [55]

The answer is C, because the water molecules are evaporating and the salt molecules are staying the same.

8 0

2 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

If a town was 90 miles away and you travel at 45 mph how long would it take to get there

notka56 [123]

Answer:

t = 2 hours

Explanation:

Given that,

Distance of the town, d = 90 miles

Speed, v = 45 mph

We need to find the time to get there. The speed of an object is given by :

$v=\dfrac{d}{t}$

Where

t is time

$t=\dfrac{d}{v}\\\\t=\dfrac{90}{45}\\\\t=2\ h$

So, the required time is 2 hours.

5 0

3 years ago

ILL MARK BRAINLIEST PLEASE HELP 2.Write the formula associating energy with power ​

ILL MARK BRAINLIEST PLEASE HELP 2.Write the formula associating energy with power