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spayn [35]
3 years ago
8

12) Which is MOST LIKELY a solid at room temperature?

Physics
1 answer:
ella [17]3 years ago
4 0
Gold
Silver
Iron
Platinum
Goes on and on 


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A motor exerts a force of 12,00N to lift an elevator 8.0m in 7.0secs what is the power produced by the motor?
SashulF [63]

Answer:

1371.4watt

Explanation:

from power=energy/time

BUT energy=force times distance

6 0
2 years ago
what is the property of an object due to its mass by which it resists any change in its position unless overcome by force ?
UkoKoshka [18]

Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

hope this helps

5 0
2 years ago
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
Aleksandr-060686 [28]

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

7 0
3 years ago
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

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

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

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

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
3 years ago
Read 2 more answers
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block\left ( m_1\right )

\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

where v_0 is the final velocity

3 0
3 years ago
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