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Ira Lisetskai [31]
3 years ago
9

A 1.75 kg stone falls from the top of a cliff and strikes the ground at a speed of 58.8 m/s. What is the height of the cliff??

Physics
1 answer:
Reil [10]3 years ago
6 0

Answer:

172.9m

Explanation:

h = 1/2 gt^

First calculat for t using

v = gt

t = v/g = 58.8/10

= 5.88secs

now h = 1/2 x 10 x 5.88^2

h =1/2 x 10 x 34.57

= 345.74/2

= 172.9m

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A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0
3 years ago
Read 2 more answers
PLEASE HELP
blagie [28]
Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg
8 0
3 years ago
Two 10cm diameter metal disks separated by a 0.63mm thick piece of pyrex glass are charged to a potential difference of 1000V. D
inessss [21]
Parallel-plate capacitor has there fore formula is

<span>C=(<span>ϵ0</span>A)/d
putting values</span>C=(8.85*10^-12*pi*.05^2)/.00063
=1.1*10^-10F then Q=CV=1.1*10^-10*1000=1.1*10^-7C 
as
<span>η=Q/A</span><span>therefore
(1.1*10^-7)/(pi*.05^2)
=1.4*10^-5C/m^ our answer
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5 0
3 years ago
The angle of reflection is the angle the ____ to the reflecting surface.
Arlecino [84]

Answer:

The angle of reflection is the angle the reflected rays make with a perpendicular line to the reflecting surface.  

Explanation:

Reflection It is the change of direction suffered by a luminous ray when hitting the surface of an object. The angle of reflection is that which is formed by the reflected ray and the normal vector to the study surface

7 0
3 years ago
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
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