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2 years ago

[10 POINTS ❗️❗️ and brainlist :)]

Physics

2 answers:

ludmilkaskok [199]2 years ago

6 0

The answer to this is D

a_sh-v [17]2 years ago

4 0

Answer:

D) Strong gravitational force of the earth

Explanation:

Gravitational force is the force that attracts to objects with mass.This forces tries to pull masses together, that is why attractive in nature.

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What are gears and what is a lever ?

blagie [28]

Answer:

agear lever or agear stick is the lever that you use to change gear in acar or other vehicle

5 0

2 years ago

Typical Pressurized Water Reactors can produce 1100 to 1500

lozanna [386]

Answer: about 1,100,000,000 to 1,500,000,000 Joules/second

Explanation:

1 MW (megawatt) = 1,000,000.00 J/s (joules per second)

1100(1,000,000) = 1,100,000,000

1500(1,000,000) = 1,500,000,000

3 0

2 years ago

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 43

kiruha [24]

Answer:

The magnification is $m = 0.3674$

Explanation:

From the question we are told that

The power of the lens is $P = -4.00 D(dioptre)$

Generally $1 dioptre = 1 \ meter$

The object distance is $u = -43 \ cm$ the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )

Generally the focal length is mathematically represented as

$f = \frac{1}{P}$

=> $f = \frac{1}{4.00 }$

=> $f = 0.25 \ m$

converting to cm

=> $f = 0.25 \ m = 0.25 * 100 = 25 \ cm$

Generally from lens equation we have that

$\frac{1}{f} +\frac{1}{v} -\frac{1}{u}$

=> $\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}$

=> $v = -15.8 \ cm$

Generally the magnification is mathematically represented as

$m = \frac{v}{u}$

=> $m = \frac{- 15.8}{-43}$

=> $m = 0.3674$

6 0

2 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

2 years ago

To make the next delivery, Sam-I-Am goes 45 mi/h for 2 hours. How far did he travel in that time?

andrezito [222]

Hello =D

This problem is about cinematic

V = 45 mi/h

t = 2 h

Then

V= X/t

X = V*t

Then

X = (45)*(2)

X = 90 mi

Best regards

5 0

2 years ago