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3 years ago

Compute the resistance in ohms of a silver block 10 cm long and 0.10 cm2 in cross-sectional area. ( = 1.63 x 10-6 ohm-cm)

1 answer:

3 0

The resistance of the silver block is given by
$R= \frac{\rho L}{A}$
where
$\rho=1.63 \cdot 10^{-6} \Omega \cdot cm$ is the silver resistivity
$L=10 cm$ is the length of the block
$A=0.10 cm^2$ is the cross-sectional area of the block

If we plug the data into the equation, we find the resistance of the silver block:
$R= \frac{(1.63 \cdot 10^{-6} \Omega \cdot cm)(10 cm)}{0.10 cm^2}=1.63 \cdot 10^{-4} \Omega$

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1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

3 years ago

A car has an acceleration of 1.5 m/s2. A net force of 2100 N is acting on the car. What is the mass of the car?

gulaghasi [49]

Answer:

Given

acceleration (a) =1.5ms2

Force(F) =2100N

R. t. c mass (m) =?

Form

F=ma(divided by m both sides)

m=F/a

m=2100/105

m=1400kg

mass of car =1400kg

8 0

3 years ago

THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period

tiny-mole [99]

Answer:

2.4583 ± 0.0207 seconds

Explanation:

The time period of a pendulum is approximately given by the formula ...

T = 2π√(L/g)

The maximum period will be achieved when length is longest and gravity is smallest:

Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds

The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:

Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds

If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.

T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2

T ≈ 2.4583 ± 0.0207 . . . seconds

We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.

5 0

3 years ago

Bobo, the clown, wants to set the world record for the furthest range traveled by a clown shot out of a cannon. Write a sentence

Furkat [3]

Solution:

The world record for the longest flight by a human cannonball is approximately 200 feet and therefore the human cannon catapult that Bobo would need is one that would be able to propel him with an initial velocity of more than 120 kilometer per hour (more than the speed of the current record holder) with the cannon directed at an angle 45 degrees above the horizonal

Learn more about the projectile motion here:

brainly.com/question/24240176

5 0

3 years ago

A boat traveling at a velocity of 20 m/s leaves island heading east . The boat slows to rest at a rate of 1.5 m/s2 How far away

ioda

The boat's initial velocity is:

$v_0 = 20 m/s$

While the boat's acceleration is

$a=-1.5 m/s^2$

with a negative sign, since the boat is slowing down, so it is a deceleration. The distance traveled by the boat until it comes to a stop can be found by using the following equation:

$v_f^2 -v_0^2 = 2aS$

where vf=0 is the final velocity of the boat and S is the distance covered. Re-arranging the formula, we can find S:

$S=\frac{-v_0^2}{2a}=\frac{-(20 m/s)^2}{2(-1.5 m/s^2)}=133.3 m$

7 0

3 years ago