Find the resistance at 100°C of a silver wire 1 mm in diameter and 1000 centimetre long ?

According to Ohm's law what would happen if the current was increased and the resistance was held constant?

Whitepunk [10]

<span>Ohm's law deals with the relation between voltage and current in an ideal conductor. It states that: Potential difference across a conductor is proportional to the current that pass through it. It is expressed as V=IR. Therefore, the correct answer is option B, the voltage would increase as well.</span>

6 0

3 years ago

What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

$g=\frac{GM}{(R+h)^2}$

where:

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$R=6.37\cdot 10^6 m$ is the Earth's radius

Here,

$h=6.38\cdot 10^6 m$

So the acceleration due to gravity is

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2$

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

$F=mg=(70)(2.45)=171.5 N$

5 0

3 years ago

One characteristic of mass wasting processes is that they a. only operate on steep slopes. b. move materials very slowly. c. ope

lord [1]

Answer:

One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.

4 0

4 years ago

If the period of a wave is 0.05 s, then what is its frequency?

yKpoI14uk [10]

Frequency = 1/time period = 1/0.05 = 20s^-1.

8 0

3 years ago

Read 2 more answers

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him

nataly862011 [7]

Answer:

a) $u_c=0\ m.s^{-1}$ & $m_c.v_c=m_b.v_b\times \cos\theta$

b) $v_c=0.0566\ m.s^{-1}$

c) $p_e=2.9218\ kg.m.s^{-1}$

Explanation:

Given:

mass of the book, $m_b=1.35\ kg$

combined mass of the student and the skateboard, $m_c=97\ kg$

initial velocity of the book, $v_b=4.61\ m.s^{-1}$

angle of projection of the book from the horizontal, $\theta=28^{\circ}$

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

$u_c=0\ m.s^{-1}$

where:

$u_c=$ initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

$m_c.v_c=m_b.v_b\times \cos\theta$

where:

$v_c=$ final velcotiy of the student after throwing the book

b)

$m_c.v_c=m_b.v_b\times \cos\theta$

$97\times v_c=1.35\times 4.61\cos28$

$v_c=0.0566\ m.s^{-1}$

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

$p_e=m_b.v_b\sin\theta$

$p_e=1.35\times 4.61\times \sin28^{\circ}$

$p_e=2.9218\ kg.m.s^{-1}$

6 0

3 years ago