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shepuryov [24]
3 years ago
14

Maria drove to the store and then to work.

Physics
1 answer:
natali 33 [55]3 years ago
8 0

Answer: C - 60km !!!

Explanation:

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When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If
Alex787 [66]

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

6 0
3 years ago
A parallel-plate capacitor is connected to a battery. The space between the two plates is empty. If the separation between the c
Maru [420]

Answer:

The final stored energy will become half.

Explanation:

 We know that stored energy in the capacitor is given as

E=\dfrac{1}{2}CV^2

C=capacitance

V=Voltage difference

E=Energy

C=\dfrac{\varepsilon A}{d}

d=Distance between plates

A=Area

E=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d}\times V^2

If the distance between plates get double ,say d' = 2 d

Then stored energy

E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{d'}\times V^2

E'=\dfrac{1}{2}\times \dfrac{\varepsilon A}{2d}\times V^2

E'=\dfrac {E}{2}

Therefore the final stored energy will become half.

5 0
4 years ago
Why does charging by rubbing happens best in dry weather?
navik [9.2K]
Maybe friction..? my best answer.
7 0
4 years ago
A 108 kg clock initially at rest on a horizontal floor requires a 639 N horizontal force to set it in motion. After the clock is
labwork [276]

Answer:

\mu_s=0.60

Explanation:

It is given that,

Mass of the clock, m = 108 kg

Force acting on it when it is in motion, F=639\ N

After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity, F' = 521 N

It is assumed to find the coefficient of between the clock and the floor. The force of friction is given by :

F=\mu_smg

\mu_s=\dfrac{F}{mg}

\mu_s=\dfrac{639\ N}{108\ kg\times 9.8\ m/s^2}

\mu_s=0.60

So, the coefficient of static friction between the clock and the floor is 0.6. Hence, this is the required solution.

5 0
3 years ago
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect
sergejj [24]

Answer:

Weight\ loss=1.6321N

Explanation:

From the question we are told that:

Weight W=85.9kg

Altitude h= 6.33 km

Let

Radius of Earth r=6380km

Gravity g=9.8m/s^2

Generally the equation for Gravity at altitude is mathematically given by

 g_s=9.8(\frac{6380}{6380+6.33})^2

 g_s=9.781m/s^2

Therefore

Weight at sea level

 W_s=9.8*85.9

 W_s=841.82N

Weight at 6.33 altitude

 W_a=9.781*85.9

 W_a=840.2N

Therefore

 Weight loss=W_s-W_b

 Weight loss=841.82-840.2

 Weight loss=1.6321N

3 0
3 years ago
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