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4 years ago

5

A 0.50 kg toy car moves at constant acceleration of 2.3 m/s2 . Determine the net applied force that is responsible for that acce

leration.

1 answer:

S_A_V [24]4 years ago

7 0

Answer:

1.15 N

Explanation:

F=ma

0.50 x 2.3 = 1.15

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A cannon is fired with an initial horizontal velocity of 20 m/s and initial vertical velocity of 25 m/s. After 3s in the air, th

Vladimir [108]

Answer:

60 m

Explanation:

After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.

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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.

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3 years ago

What are the strengths and weaknesses of the four methods of waste management?<br>

Nitella [24]

Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;

* It creates employment

* It keeps the environment clean

* The practice is highly lucrative

* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

* The process is mostly

* There is a need for global buy-in

* The resultant product had a short life

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

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4 years ago

A particle of mass 2kg resting on a smooth table attached to a fixed point on the table by a rope 1.0m making 300revolution per

zepelin [54]

Answer is: 1973.17N aprox.
step by step in the pic below

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3 years ago

A horizontal force of 100 N is required to push a 50 kg crate across a factory floor at a constant speed. What is the accelerati

luda_lava [24]

Answer:

a = 2m/s^2

Explanation:

Force (F) = 100 N

Mass (m) = 50 kg

Here,

F = m×a

100 = 50 × a

a = 100÷50

a = 2m/s^2

Thus, the acceleration on the cart is a = 2m/s^2

-TheUnknownScientist

6 0

3 years ago