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2 years ago

In the formula vy = ayt, what is the value of ay for an object in projectile motion?

Physics

2 answers:

Sergeu [11.5K]2 years ago

8 0

–9.8 m/s<span>2

Just took it and got it right!</span>

Deffense [45]2 years ago

6 0

Answer : The correct option is (A) $-9.8\ m/s^2$

Explanation :

The formula is given as :

$v_y=a_yt$

Where,

$v_y$ is vertical component of velocity.

$a_y$ is the vertical component of acceleration.

In a projectile motion, the particle moves in a path under the influence of gravity.

So, the vertical component of acceleration is -g i.e acceleration due to gravity.

Hence, the correct option is (a) " $a_y=-9.8\ m/s^2$ ".

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Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

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$v = \sqrt{\frac{GM}{(r + h)}}$

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$M = 4.54 \times 10^{23} kg$

Now the density is given as

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2 years ago

4- by eating food only will you get energy?

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Answer:

yes

Explanation:

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2 years ago

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

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Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

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