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Anuta_ua [19.1K]

3 years ago

10

A block is hung by two ropes angled at 30º and 60º respectively. What is the ratio of the tension in the second rope to the tens

ion in the first rope?

1 answer:

WINSTONCH [101]3 years ago

8 0

The ratio between 30° and 60° is 1:2

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A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra

Tatiana [17]

Answer:

The centripetal acceleration of the car is $8\ m/s^2$ .

Explanation:

Let the mass of the car, $m=10^3\ kg$

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(20\ m/s)^2}{50\ m}$

$a=8\ m/s^2$

So, the centripetal acceleration of the car is $8\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

Which is an example of transfroming potential energy to kinetic

Rudik [331]

Answer:

A ball being dropped to the ground

6 0

3 years ago

Read 2 more answers

Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster

astraxan [27]

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s$

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

7 0

3 years ago

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i

slega [8]

Answer:

The appropriate solution is:

(a) $\frac{1}{4}(I_o)$

(b) $\frac{1}{4} (u_o)$

(c) $\frac{1}{2}B_o$

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒ $\frac{I}{I_o} =\frac{r_o_2}{r_2}$

On applying cross-multiplication, we get

⇒ $I=I_o\times \frac{1_2}{2^2}$

⇒ $=I_o\times \frac{1}{4}$

⇒ $=\frac{1}{4} (I_o)$

(b)

As we know,

⇒ $\frac{u}{u_o} =\frac{I}{I_o}$

By putting the values, we get

⇒ $u=\frac{1}{4}(u_o)$

(c)

⇒ $\frac{B^2}{B_o^2} =\frac{u}{u_o}$

$=\frac{I}{I_o}$

⇒ $B=B_o\times \sqrt{\frac{1}{4} }$

⇒ $=\frac{1}{2}(B_o)$

4 0

3 years ago

A 174 pound Jimmy Cheek is riding on a 54 ft diameter Ferris Wheel. The normal force on Jimmy Cheek is 146 pounds when Jimmy is

Veronika [31]

To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.

I will also attach a free body diagram that allows a better understanding of the problem.

For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore

$F_c = W-N$

$m\omega^2r = W-N$

Here,

m = Net mass

$\omega$ = Angular velocity

r = Radius

W = Weight

N = Normal Force

$m\omega^2r = 174-146$

The net mass is equivalent to

$F = mg \rightarrow m = \frac{F}{g}$

Then,

$m = \frac{174lb}{32.17ft/s^2}$

Replacing we have then,

$(\frac{174lb}{32.17ft/s^2})\omega^2 (54ft) =174lb-146lb$

Solving to find the angular velocity we have,

$\omega = 0.309rad/s$

Therefore the angular velocity is 0.309rad/s

6 0

3 years ago