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3 years ago

If we decrease the amount of force, and kelp all other factors the same, what will happen to the amount of work?

Physics

1 answer:

Artist 52 [7]3 years ago

8 0

Answer:

Distance will decrease and work will decrease:

F = m a Newton's Second Law

a = F / m decreasing force will decrease acceleration

S = 1/2 a t^2 = 1/2 (F / m) t^2 distance traveled will decrease as force decreases

W = F * S work will decrease as both force and distance decrease

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Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which

GuDViN [60]

Answer:

i. 0.34

ii. 0.4

iii. 1700 w/m²

iv. 2211.36 w/m²

Explanation:

Given that

Irradiation of the plate, G = 2500 w/m²

Reflected rays, p = 500 w/m²

Emissive power, E = 1200 w/m²

See attachment for calculations

8 0

4 years ago

Cheetahs the worlds fastest land animals can run up to about 125km/h. a cheetah chasing an impala runs 32m [N], then suddenly tu

AleksAgata [21]

Av Speed = total distance / time time = 32+ 46 / 2.7 = 28 m/sec
Av velocity = total displacement / time total = S / t
S = sqrt( 32^2 +46^2) = 56 m
Av Velocity = 56/ 2,7 = 20.75 m/sec
with angle tan^-1 = 0.7 north west ( about 35 degrees north west)

3 0

3 years ago

when the particles of the medium move back and forth along the direction of the wave motion, the wave is a

Bezzdna [24]

Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.

8 0

3 years ago

Read 2 more answers

The speed of sound is measured to be 340 m/s on a certain day.

ziro4ka [17]

Answer: 1,224 km/h

Explanation:

To do this, we pick the first unit and convert
Picking m first and converting to km:
Since we're converting from a non-prefix to a prefix, we divide the value by the prefix were taking it to. In this case, kilo = 10³ which means we're going to divide our value by 1000 to convert it from m to km
340 m/s ÷ 1000 = 0.34 km/s
Now, let's convert our seconds to hour:
We'll need to calculate how many hours is equivalent to one second first;
1 hr = 60×60 seconds
X hr = 1 second
*Cross multiply*
1 × 1 = X × 60 × 60
1 = 3,600 X
X = 1 / 3,600
X = 2.778×10⁻⁴ hour
So, in the place of "1 Second", we're going to be inserting 2.778×10⁻⁴ hour instead
0.34 km / s = 0.34 km / 2.778×10⁻⁴ hour
(0.34 / 2.778×10⁻⁴) km/hour
1,224 km/h.
340 m/s = 1,224 km/h

6 0

2 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago