A student draws a distance verses time graph to describe the motion of a model train. What does a

An astronaut is moving in space when a big explosion occurs about 50 meters behind him. How will the astronaut come to know abou

LUCKY_DIMON [66]

Answer:

The correct answer is B.

The astronaut will know due to the light from the explosion.

Explanation:

Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.

Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.

Cheers!

7 0

3 years ago

You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is

scZoUnD [109]

Answer:4.34 miles

Explanation:

first Elevation = $19^{\circ}$

After 1 minute Elevation changes to $59^{\circ}$

Ditsance travelled in 1 minute = $600\times \frac{1}{60}$ =10 mile

Now

tan59= $\frac{H}{x}$

H=xtan59

tan19= $\frac{H}{10+x}$

H= $\left ( 10+x\right )tan19$

Equating H

we get

1.319x=10tan19

x=2.61 miles

H= $2.61\times tan59$ =4.34 miles

4 0

3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$