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3 years ago

6

Which type of cloud would you expect to be involved in some form of precipitation?

1 answer:

fomenos3 years ago

6 0

Stratus clouds maybe

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What likely happens to the hart rate as the cardiac mucle gets stronger

Kaylis [27]

The heart rate will likely decrease. As the cardiac muscle, or heart, gets stronger, it takes less effort to pump more blood. As a result, the heart will probably beat less, decreasing the heart rate. This is why athletes often have lower heart rates than the average person.

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3 years ago

A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are

Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, $\theta=30^{\circ}$

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

$W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)$

$v = 0$

$Fd\ cos\theta=\dfrac{1}{2}m(u^2) F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N$

The force required to push to stop the car is 288.67 N

3 0

3 years ago

What is the mass of a football in free fall with a total force due to gravity of 782N?

Mama L [17]

5

Explanation:

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4 0

3 years ago

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An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}$

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

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3 years ago

Atoms combine when _____.

Viefleur [7K]

I think the answer is b

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3 years ago

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