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3 years ago

Help fast please!!!!!

Physics

2 answers:

Svet_ta [14]3 years ago

8 0

Answer:

the answer is A

Hope tis helps:)

Explanation:

inysia [295]3 years ago

6 0

We're told that the planets have EQUAL MASS.

If that's true, then the strength of the gravitational forces between
each planet and the star depends only on the distance between
them ... the farther a planet is from the star, the smaller the
gravitational forces are IF we're talking about planets with
equal masses.

Planet-X is closer to the star, and Planet-Y is farther from it.
From this we know that the gravitational forces between the
star and Planet-X are greater, and the forces between the star
and Planet-Y are smaller.

'A' says this.

'B' is totally absurd, because it talks about gravity repelling things.

'C' says exactly the opposite for the two planets.

'D' says that distance doesn't matter. We know this is absurd,
simply because we're never pulled toward Jupiter in our daily life.

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The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g

monitta

Answer:

55.56kg

Explanation:

Given:

F= 52N

a=0.936m/s²

Applyinc Newton's second law, that states: force is equal to mass times acceleration.

F = ma

m=F/a =>52 / 0.936

m=55.56kg

5 0

3 years ago

Which statement is true regarding flexibility? A. joint's range of motion cannot be maintained with age. b.A joint's range of mo

Alex777 [14]

<span>the statement that is true regarding flexibility is : b. a joint's range of motion will be lost if the joint is not used regularly. Our body is like a machine. If we not constantly heat it up, our body will be more prone to injury. We can see that the old people who lived within the tribe in the middle of the mountain are far stronger than the one who lived in the city.</span>

3 0

3 years ago

Read 2 more answers

A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L

⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒ P = 24624000 N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax we use σmax

⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

7 0

3 years ago

The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo

notka56 [123]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t = 0.00075 s

The impulsive force is given by the formula:

$F=\frac{mv}{t}$

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

$F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N$

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

Learn more here: brainly.com/question/25892144

4 0

2 years ago