Ask question

Ask question

All categories

3 years ago

6

Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How f

ar behind the cart will the ball land, compared to the distance in the original experiment?
(A) half as far
(B)twice as far
(C) four times as far
(D) the same distance
(E) by a factor not listed above

1 answer:

expeople1 [14]3 years ago

5 0

Answer:

(C) four times as far

Explanation:

As we know that the range of the launcher is given as

$R = \frac{v^2sin(2\theta)}{g}$

here we know that all the parameters will remain same but only change is the velocity is doubled

So we will have

$R' = \frac{(2v)^2 sin2\theta}{g}$

so we have

$R' = 4 R$

so here correct answer will be

(C) four times as far

You might be interested in

In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will

Kamila [148]

Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

5 0

2 years ago

Why does the area around the equator stay about the same temperature year round?

Verdich [7]

Axial Tilt and Sun Energy

This axial tilt means that during the Earth's journey around the sun the poles receive varying amounts of sunlight. The equator, however, receives relatively consistent sunlight all year. The consistency of energy means the equator's temperature stays relatively constant all year.

3 0

3 years ago

Which energy transformation occurs when a butane lighter is lit?

UNO [17]

It is electrical energy into connected energy has in relation to the question

7 0

2 years ago

Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an

galben [10]

Answer:

$\triangle P=1.95*10^{-4}$

Explanation:

Mass $m=0.001$

Diameter $d=1.2m$

Length $l=10m$

Generally the equation for Volume flow rate is mathematically given by

$Q=AV$

$V=\frac{Q}{\pi/4D^2}$

$V=\frac{0.001}{\pi/4(1.2)^2}$

$V=8.84*10^{-4}$

Generally the equation for Friction factor is mathematically given by

$F=\frac{64}{Re}$

Where Re

Re=Reynolds Number

$Re=\frac{pVD}{\mu}$

$Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}$

$Re=1040$

Therefore

$F=\frac{64}{Re}$

$F=\frac{64}{1040}$

$F=0.06$

Generally the equation for Friction factor is mathematically given by

$Head loss=\frac{fLv^2}{2dg}$

$H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}$

$H=19.9*10^{-9}$

Where

$H=\frac{\triangle P}{\rho g}$

$\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}$

$\triangle P=H*\rho g$

$\triangle P=1.95*10^{-4}$

6 0

3 years ago

Which of the following would decrease the magnetic field around a wire?

shusha [124]

The correct answer is
<span>A. decreasing the current

In fact, the magnetic field produced by a current carrying wire is given by
</span> $B(r) = \frac{\mu_0 I}{2 \pi r}$
<span>where
</span> $\mu_0$ is the vacuum permeability
<span>I is the current in the wire
r is the distance from the wire at which the field is calculated

We see from the formula that the intensity of the field, B, is directly proportional to the current I, so if the current decreases, the magnetic field strength B decreases as well.</span>

4 0

3 years ago