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slamgirl [31]
3 years ago
7

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of

the incline, and its speed after it has traveled 6.00 mm to the bottom of the incline is 3.80 m/s.What is the speed of the block when it is 3.00 m from the top of the incline?
Physics
1 answer:
ehidna [41]3 years ago
5 0

Answer:

Explanation:

Given

Speed of block at bottom is v=3.8\ m/s

Distance traveled s=6\ m

initial velocity is zero

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

(3.8)^2-0=2\times a\times 6

a=1.203\ m/s^2

when it is 3 m from top then

v^2-u^2=2as

v^2-0=2\times 1.203\times 3

v=2.68\ m/s      

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4 0
2 years ago
A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha
denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

E = - \frac{N*\Delta \phi}{\Delta t}  

<u>Where:</u>

<em>N = is the number of turns = 1   </em>

<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T                     </em>

Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0
4 years ago
From the top of a cliff, a person tosses a pebble straight downward with an initial velocity of -9.0 meters/second. After 0.50 s
Irina-Kira [14]
Y - yo = Vo*t - g * (t^2) / 2

Vo = - 9.0 m/s
t = 0.50 s

=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.

Answer: option c) - 5.7
8 0
3 years ago
While playing Quidditch you throw the quaffle straight down to the person below you at
oee [108]

Answer:

the ans will be because it has 1.672

3 0
4 years ago
The radius of the wheel is 40cm and the radius of the
snow_tiger [21]

Answer:

M.A = load/ effort

1200N/400N

= 3

velocity ratio= radius of wheel/radius of Axle

40cm/10cm

=4

efficiency= 3/4*100

75%

4 0
3 years ago
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