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3 years ago

A spring of force constant 1500 Nm-l is acted

Physics

1 answer:

olchik [2.2K]3 years ago

7 0

Answer:

1.876 J

Explanation:

First, let’s calculate the compression of the spring from the Hooke’s law:

F=kx,

here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.

Then, we get:

x=F/k=(75 N)/(1500 N/m)=0.05 m.

Finally, we can find the potential energy stored in the spring:

PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.

correct my answer if it's wrong ^^

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4. If you swing wide to the left before turning right, another driver may try to pass you on the right. True or False

olchik [2.2K]

Answer:

True.

Explanation:

Don't turn wide to the left as you start the turn. A driver behind may think you are turning left and try to pass you on the right. You may crash into the other vehicle as you complete your turn.

Instead, slowly give yourself and others more time to avoid problems, keep the rear of the vehicle close to the curb. This will stop other drivers from passing you on the right. This is called (button Hook)

If you are driving a truck or bus that cannot make the right turn without swinging into the other lane, turn wide as you complete the turn.

8 0

2 years ago

1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of

lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0

3 years ago

Stars of spectral type A and F are considered ________.

LekaFEV [45]

Answer:

B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.

Explanation:

The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:

• They live at least a few billion years, allowing life a chance to evolve. More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.

• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.

• They emit sufficient radiation at wavelengths conducive to photosynthesis.

• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.

Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.

4 0

2 years ago

An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

$Tsin\theta= \frac{mv^2}{r}$

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

$Tcos\theta = mg$

Matching both expressions with respect to the tension we will have to

$T = \frac{\frac{mv^2}{r}}{sin\theta}$

$T = \frac{mg}{cos\theta}$

Then we have that,

$\frac{\frac{mv^2}{r}}{sin\theta} = \frac{mg}{cos\theta}$

$\frac{mv^2}{r} = mg tan\theta$

Rearranging to find the velocity we have that

$v = \sqrt{grTan\theta}$

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is

$r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}$

$r = \frac{11.1}{2}+2.41$

$r = 7.96m$

Replacing we have that

$v = \sqrt{(9.8)(7.96)tan(14.5\°)}$

$v = 4.492m/s$

Therefore the speed of each seat is 4.492m/s

6 0

3 years ago

A 2.51 kg ball is attached to a ceiling by a 1.19 m long string. The height of the room is 3.45 m. The acceleration of gravity i

ANTONII [103]

Answer:

55,42 J

Explanation:

Since the height of the room is 3.45 m (distance between the floor and the ceiling) the difference between this value and the length of the rope 1.19 m; it will be equal to (3.45-1.19) =2.26 m. If we take as a reference point (Ep=0) the floor of the room, then the potential energy will be equal to Ep = M * g * h, replacing values in this equation (2.5 kg * 9.81 m/s2 * 2.26 m) will be 55,42 (N * m) or Jules.

6 0

2 years ago