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bagirrra123 [75]

2 years ago

12

If your weight is 588N on the earth, how far should you go from the centre of the earth so that your weight will be 300N? The ra

dius of the earth is 6400km and the value of g on the earth surface is 9.8m/s2.

1 answer:

Rom4ik [11]2 years ago

5 0

Answer:

Hmm, Well I am pretty sure the answer is 288N

Explanation:

I got the answer 288N but I am not sure if that is correct. Hope it helps! :D

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g Consider a (12.5 A) cm long metal bar moving horizontally across a vertical magnetic field at a speed of (2.40 B) m/s. The mag

Misha Larkins [42]

Answer:

Explanation:

emf due to movement of a rod of length L in a perpendicular to magnetic field B with velocity v is given as

emf = BLv

Putting in the given values ,

E = 2.45 x 12.5 x 10⁻² x 2.4

= .735 V

This emf produces current in resistance . Power consumed by resistance

V² / R where R is resistance , V is emf induced .

Power = .735² / 1.2

= .45 W .

7 0

2 years ago

Which of the following observations represent conclusive evidence of an interaction? (Select all that apply.)

9966 [12]

Answer:

a,b,d and e are correct.

Explanation:

a) Change of temperature without change of velocity is a conclusive evidence of an interaction. As the temperature of a body will only change if heat energy is flows in or out of the body with surrounding. So, option a) is correct.

b) Change of the direction without change of speed is an evidence of an interaction. The direction changes when the object is under a perpendicular acceleration acceleration which clearly means the particle is interacting or external force applied. So, option b is correct.

d) Change of shape or configuration without change of velocity is conclusive evidence of an interaction. The shape can change only if external stress is applied on the particle. So, Option d is correct.

e) Change of identity without change of velocity is a conclusive evidence of an interaction. The identity of an object can only change if it interacts with surroundings. So, option e is correct

Therefore the only incorrect option is c .

6 0

3 years ago

A glass falls from a table top and smashes on the floor 0.6 seconds later. How high is the table?

soldi70 [24.7K]

Answer:

1.8 m

Explanation:

Given: Glass falls from a table, smashes 0.6 seconds later

To find: How high a table is

Formula: Vv=gt, dv=1/2gt^2, t=2d/g

Solution: A table's <em>height</em> is measured from the top of the edge down to the floor. The tables are shown both have a height of 30 inches, which is common for many tables.

<u>Data</u>

t = 0.6s

g = 9.81 m/s²

d = ?

<u>Equation</u>

d = $\frac{1}{2}$ $gt$ ²

<u>Math & Units</u>

d = 4.905 (0.6²)
d = 442.676

Hence the table is 1.8 m high

5 0

2 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

2 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago