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2 years ago

3

An object moves with a constant speed of 30m/s on a circular track of radius 150m, What is the acceleration of the object? A) 0m

/s2 b) 0.17m/s2 c) 5.0 m/s2 d) 6.0 m/s2

1 answer:

Bumek [7]2 years ago

7 0

Answer:

C

Explanation:

read the question and they mean to mutilpy toea 30 m and subterc

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago

Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0

2 years ago

A 3.0-kilogram object is acted upon by an impulse having a magnitude of 15 newton•seconds. What is the magnitude of the object’s

Ilia_Sergeevich [38]

The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.

3 0

2 years ago

A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

VMariaS [17]

Answer:

Hence, work done= 287.54 J

Explanation:

Given data:

angle of ramp with the ground θ =20°

force applied = 76 N

work done on the crate to slide down 4 m down the ramp

W= F×d cosθ ( only the cos component of the force will slide the crate down)

W= 76×4×cos20= 287.54 J

4 0

2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

$g=10 m/s^2$ is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

$p=\frac{F}{A}$

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

6 0

2 years ago