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3 years ago

A ball is thrown horizontally. What is the ball's acceleration at its highest point

1 answer:

4 0

At the highest point of motion the ball comes to rest momentarily,but it is being pulled down due to the effect of gravity,so its net acceleration is downwards. So,just after that point,it starts falling downwards.

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What will occur when the trough of wave A overlaps the trough of wave B?

densk [106]

Constructive interference will occur, which means the waves will combine.

In destructive inference, the waves cancel each other out.

Hope this helps :)

3 0

3 years ago

Read 2 more answers

An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri

Anettt [7]

Answer:

270 mi/h

Explanation:

Given that,

To the south,

v₁ = 300 mi/h, t₁ = 2 h

We can find distance, d₁

$d_1=v_1\times t_1\\\\d_1=300\times 2\\\\d_1=600\ \text{miles}$

To the north,

v₂ = 250 mi/h, d₂ = 750 miles

We can find time, t₂

$t_2=\dfrac{d_2}{v_2}\\\\t_2=\dfrac{750\ \text{miles}}{250\ \text{mi/h}}\\\\t_2=3\ h$

Now,

Average speed = total distance/total time

$V=\dfrac{d_1+d_2}{t_1+t_2}\\\\V=\dfrac{600+750}{2+3}\\\\V=270\ \text{mi/h}$

Hence, the average speed for the trip is 270 mi/h.

3 0

3 years ago

Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

3 0

4 years ago

“How bright would the sun appear to an observer on Earth if the sun were four times farther from Earth than it actually is? Expr

Alenkasestr [34]

Like a lot of other things, (gravity, sound, electrostatic force), brightness also decreases as the square of the distance.

When the source moves to a new position that's 4 times as far away, its apparent brightness becomes (1/4^2) its original value.

That's 1/16 .

4 0

4 years ago

Two people are standing on rollerskates. One is more massive than the other. They push against each other and move away. How do

Gala2k [10]

If they push away, C. The more massive person moves slower.

6 0

3 years ago