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puteri [66]
3 years ago
14

A cart is pushed on a frictionless floor with 29.6 Newtons of force along a distance of 5.9 meters in 8.1 seconds. How much powe

r is exerted
during this time period?
Physics
1 answer:
marta [7]3 years ago
8 0

Answer:

Power = 21.6 Watts

Explanation:

Power = \frac{W}{t}

Where W = work and t = time.

W = Fs

Where F is force and s is displacement.

29.6N(5.9m) = 174.64J

W = 174.64J

Power = \frac{174.64J}{8.1s}

Power = 21.6 Watts

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A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of
Alik [6]

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

f=\frac{100}{1.75}=\frac{400}{7}

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm

∴ The new near point is 17.4 cm

7 0
3 years ago
SO! I'VE GOT A COUPLE QUESTIONS RELATING TO SCIENCE (WHERE ARE MY SCIENCE WHIZZES?) TRUST ME THERE WILL BE MORE
Helga [31]

Answer:

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Explanation:

5 0
3 years ago
Read 2 more answers
A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t
navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that v^2=u^2+2as

16^2=24^2+2\times a\times 50

a=-3.2m/sec^2

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So v^2=u^2+2as

0^2=24^2-2\times 3.2\times s

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration a=-3.2m/sec^2

So time t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec

7 0
3 years ago
Displacement vectors of 4 km north, 2km south, 5 km north, and 5km south combine to a total displacement of A 2 km south B 16 km
prohojiy [21]
I would prolly say C. 16 km south.
4 0
3 years ago
Read 2 more answers
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
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