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3 years ago

Grade 10 My smart Physics people help me with this review question please

Physics

1 answer:

Kaylis [27]3 years ago

3 0

Answer: a = 1, b = 9, c = 6, d = 4

Explanation:

With no velocity, total mechanical energy is all gravity potential energy

PE = mgh = 80.0(9.81)(25) = 19,620 J = 1.96e4 J

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Which statement is true? The speed of sound in air is inversely proportional to the temperature of the air. The speed of sound i

sasho [114]

<span> the speed of sound in air is directly proportional to the temperature of the air. The speed of sound depends on the temperature of the surrounding air, this can be represented by a speed of sound in air formula: v = 331m/s + 0.6m/s/C * T (where T is temperature)</span>

6 0

3 years ago

Read 2 more answers

A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.

vekshin1

Answer:

A) If the paintball stops completely the magnitude of the change in the paintball’s momentum is $p=0.273kg*m/s$

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is $p=0.546kg*m/s$

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

A) We use the formula of momentum $p=mv$ , so we have $p=0.0032kg*85.3m/s=0.273kg*m/s$

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

$p=2*0.0032kg*85.3m/s=0.546kg*m/s$ .

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

$F=\frac{\Delta p}{\Delta t}$ , as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0

3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!