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soldier1979 [14.2K]
3 years ago
11

Pedro uses a bar magnet to pick up a nail. He then touches the tip of the nail to some staples.

Physics
2 answers:
scoundrel [369]3 years ago
7 0
This isn’t a question
allochka39001 [22]3 years ago
7 0

Answer:

B) The nail has become a temporary magnet, while the bar magnet remains a permanent magnet.

Explanation:

I hope this helped you ;)

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A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th
Liono4ka [1.6K]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, m=\dfrac{F}{a}

m=\dfrac{70\ N}{7.5\ m/s^2}

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

W=m\times g

W=9.34\ kg\times 1.62\ m/s^2

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

3 0
3 years ago
The speed of an object undergoing constant acceleration increased from 8.0 meters per second to 16.0 meters per second in 10. Se
saw5 [17]

v₀ = initial speed of the object = 8 meter/second

v = final speed of the object = 16 meter/second

t = time taken to increase the speed = 10 seconds

d = distance traveled by the object in the given time duration = ?

using the kinematics equation

d = (v + v₀) t/2

inserting the above values in the above equation

d = (16 + 8) (10)/2

d = 120 meter


6 0
3 years ago
A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of
Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0
3 years ago
Solar evaporation ponds are shallow so that the sun can heat them up, causing the water to evaporate, which increases the salt c
ehidna [41]

Answer:

B. stearothermophilus and S. ruber

Explanation:

B. stearothermophilus and S. ruber

In solar evaporation ponds the temperature is higher and the salt concentration is also higher because of the water evaporated so sunder such extreme conditions this hybrid bacteria is capable of surviving. B. stearothermophilus is thermophilus bacteria which grows at high temperature and S. ruber is halophilic bacteria which grows in saline environment. So, these two bacteria best suited for the above hybrid condition.

6 0
3 years ago
Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d
Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

h_{fg} = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

T_{L} = -18.77°C = 254.23 K

At saturation pressure  800 kPa, the saturation temperature is

T_{H} = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, q_{reject} = h_{fg} = 171.82 kJ/kg

We know COP of heat pump

COP = \frac{T_{H}}{T_{H}-T_{L}}

        = \frac{304.31}{304.31-254.23}

         = 6.076

Therefore, Work out put, W = \frac{q_{reject}}{COP}

                                              = 171.82 / 6.076

                                              = 28.27 kJ/kg

8 0
4 years ago
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