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Alisiya [41]
3 years ago
13

Anyone who answer will get brainlest

Physics
2 answers:
Ganezh [65]3 years ago
6 0

Answer:

D

Explanation:

work = Force x distance

work = 2000 x 2 = 4000

power = work÷ time

4000/.8 = 5000W

lina2011 [118]3 years ago
3 0

Answer:

The answer is D 5000w

Explanation:

work = Force x distance

work = 2000 x 2 = 4000

power = work÷ time

4000/.8 = 5000W

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What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

6 0
3 years ago
Read 2 more answers
How much would you have to shrink a baseball to a black hole?
Vesna [10]

The Schwarzchild radius is . . . <em>R = 2GM/c²  .</em>

Just like you, I'm not completely sure what that means.  But it DOES use the mass of a black hole to calculate a radius associated with it, so with 15 Brainly points at stake, it seems like a good-enough formula to use for an answer.

Before I proceed, I really should ask you whether you're talking a softball or a hardball, but again . . . . .

So R = 2GM/c²

G = the  gravitational constant = 6.67 x 10⁻¹¹ N-m²/kg²

M = mass of the baseball = 145 grams = 0.145 kg

c = speed of light = 3 x 10⁸ m/s

R = 2 (6.67 x 10⁻¹¹ m³/kg-s²) (0.145 kg) / (3 x 10⁸ m/s)²

R = (2 x 6.67 x 10⁻¹¹ x 0.145 / 9 x 10¹⁶) (m³-kg-s² / kg-s²-m²)

R = ( 1.9343 x 10⁻¹¹ / 9 x 10¹⁶) (m³-kg-s²/m²-kg-s²)

R = (0.2149 x 10⁻²⁷) meter

<em>R = 2.149 x 10⁻²⁸ meter</em>

For reference: The radius of a Hydrogen atom is 1.2 x 10⁻¹⁰ meter.

So in order to make a black hole out of a baseball, you have to crunch the baseball down to around  0.000000000000000001791 the size of a Hydrogen atom.

Would that be a problem for you ?

3 0
3 years ago
a runner makes one lap around a 200m track in 25s, what is the runners (a) average speed and (b) average velocity
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5 0
3 years ago
A bug is moving along a meter stick in the negative direction at a constant speed of 0.85cm/s. After 42 seconds have passed the
Agata [3.3K]

Given that,

bug speed, v= 0.85 m/s

time, t =42 s

Final position of bug on meter stick was 27 cm

Starting position of bug on meter stick = ?

Since we know that,

s = vt

s= 0.85*42 = 35.7 cm

this is the distance covered by bug in the given time and velocity.

since the bug is moving in negative direction, starting point will be:

27.0 cm+ 35.7 cm = 62.7 cm

The bugs starting position on meter stick was 62.7 cm.

3 0
4 years ago
One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing co
Kruka [31]

Answer:

v_{1}=1.886 \frac{m}{s}

β= 57.99 south of east

Explanation:

v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=?

Velocity in axis x the two balls come one from east and west

m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\  v_{fx1}=1 \frac{m}{s} \\

Velocity in axis y initial is zero so:

v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}

v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}

Angle is find using:

tan(β)=\frac{v_{fy}}{v_{fx}}

\beta =tan^{-1}*\frac{1.6}{1}=57.99

5 0
4 years ago
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