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Makovka662 [10]

3 years ago

11

a pistol fires a bullet towards a target located 175m away. The bullet is traveling at 320 m/s. How long does it take for the bu

llet to hit the target?

1 answer:

inn [45]3 years ago

7 0

Time = distance/speed

Time = (175 m) / (320 m/s)

<em>Time = 0.547 second</em>

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A 16-cm-long straight line connects the center of a turntable to its edge. The turntable rotates counter-clockwise at 45 rpm. A

Bond [772]

Answer:

$\mathbf{V_x = 3.25 \ cm/s}$

$\mathbf{V_y = 1.29\ cm/s}$

Explanation:

Given that:

The radius of the table r = 16 cm = 0.16 m

The angular velocity = 45 rpm

= $45 \times \dfrac{1}{60}(2 \pi)$

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

$t = \dfrac{r}{v}$

$t = \dfrac{0.16}{0.035}$

t = 4.571s

So the angle made by the radius r with the horizontal during the time the bug gets to the end is:

$\theta = \omega t$

$\theta = 4.712 \times 4.571$

$\theta = 21.54^0$

Now, the velocity components of the bug with respect to the table is:

$V_x = Vcos \theta$

$V_x = 0.035 \times cos (21.54^0)$

$V_x = 0.0325 \ m/s$

$\text {V_x = 3.25 \ cm/s}$ $\mathbf{V_x = 3.25 \ cm/s}$

Also, for the vertical component of the velocity $V_y$

$V_y = V sin \theta$

$V_y = 0.035 \times sin (21.54^0)$

$V_y = 0.0129\ m/s$

$\mathbf{V_y = 1.29\ cm/s}$

6 0

2 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

If a car accelerates uniformly from rest to 15 meters

Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0

3 years ago

Read 2 more answers

To escape from a horrible fire, two people are forced to jump from the third story of a burning building onto solid concrete. Wh

erica [24]

The bouncing person because the bounce helped him survive

8 0

3 years ago

How did harlow shapley conclude that the sun was not in the center of the milky way galaxy?

timurjin [86]

We learned that We are in the disk of the Galaxy, about 5/8 of the way from the center.

<h3>What is the work of Harlow Shapley?</h3>

Shapley, who was headquartered in Boulder, Colorado, used Cepheid variable stars to estimate the size of the Milky Way Galaxy and its position relative to the Sun. In 1953, he published his "liquid water belt" theory, today known as the concept of a livable zone.

There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.

Approximately 5/8 of the way from the galaxy's nucleus, we are in the disc. William Herschel believed that the Sun and Earth were about in the middle of the vast cluster of stars known as the Milky Way.

To learn more about Harlow Shapley's original estimate go to - brainly.com/question/28145909

#SPJ4

3 0

2 years ago