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Anna11 [10]
1 year ago
10

A 1.0 kg block is attached to an unstretched

Physics
1 answer:
KonstantinChe [14]1 year ago
4 0

Answer:

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

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The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the
Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

E = \dfrac{kq}{r^2}

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

q= \dfrac{Er^2}{k}

now,

q= \dfrac{180000\times 0.015^2}{9\times 10^9}

      q = 4.5 x 10⁻⁹ C

      q = 4.5 nC

the charge on the object is equal to 4.5 nC

8 0
2 years ago
Read 2 more answers
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
1 year ago
a 59kg physics student jumps off the back of her laser sailboat (42kg). after she jumps the laser is found to be travelling at 1
evablogger [386]

From the conservation of linear momentum of closed system,

Initial momentum = final momentum

Mass of the student, M = 59 kg

Mass of the laser boat, m = 42 kg

Initial speed of student + laser boat, u =0

Final speed of laser boat, v = 1.5 m/s

Final speed of the student = V

(M+m) u =M V +m v

0 = (59 kg) V + (42 kg) (1.5m/s)

V = - 1.06 m/s

Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.

5 0
2 years ago
Deltas, like this one, are formed at the mouth of a river. They are formed by the ___________ of sediments, soil, sand, gravel,
adelina 88 [10]
Compression is the correct answer
7 0
2 years ago
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The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the
Diano4ka-milaya [45]

Answer:

<em>The object could fall from six times the original height and still be safe</em>

Explanation:

<u>Free Falling</u>

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

U=m.g.h

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

\displaystyle K=\frac{1}{2}m.v^2

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

\displaystyle m.g.h=\frac{1}{2}m.v^2

Solving for h

\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}

If the drop had occurred in the Moon, then

\displaystyle h_M=\frac{v_M^2}{2g_M}

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}

\displaystyle h_M=6\frac{v^2}{2g}=6h

This means the object could fall from six times the original height and still be safe

6 0
2 years ago
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