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3 years ago

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A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 3240 meters away

?

2 answers:

Alik [6]3 years ago

7 0

4.5 seconds you devoured 3240/720=4.5

Nataliya [291]3 years ago

4 0

Answer:

Time taken, t = 4.5 seconds

Explanation:

Speed of the bullet, v = 720 m/s

Distance to be covered by the target, d = 3240 m

Let t is the time required for the bullet to strike a target. The time taken is calculated as :

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

$t=\dfrac{3240\ m}{720\ m/s}$

t = 4.5 seconds

So, the time required for the bullet to strike a target is 4.5 seconds. Hence, this is the required solution.

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A flashing red traffic light at an intersection means

k0ka [10]

Just do what u would do if u were at a stop sign

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4 years ago

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

7. Which quantity is the transfer of energy?<br> force<br> motion<br> work

Arisa [49]

In this exercise we have to know the definition of energy to understand what is transferred to a body, like this:

Work

<h2>What is energy?</h2>

Despite being used in many different contexts, the scientific use of the word energy has a well-defined and precise meaning: Innate potential to perform work or perform an action. Anything that is working, moving another object, or heating it up, for example, is expending (transferring) energy.

With this definition we can say that the only alternative that responds to this is work.

See more about energy at brainly.com/question/1932868

6 0

2 years ago

Read 2 more answers

How much time would it take for an airplane to reach its destination if it traveled at an average velocity of 820 km/hr NE for a

Vladimir79 [104]

Answer:

Time, t = 6.34 hours.

Explanation:

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{displacement}{time}$

Therefore, making time the subject of formula;

$Time = \frac{displacement}{velocity}$

Given the following data;

Displacement = 5200km

Average velocity = 820km/hr

Substituting into the equation, we have;

$Time = \frac{5200}{820}$

Time = 6.34 hours.

<em>Hence, it would take 6.34 hours for the airplane to reach its destination. </em>

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3 years ago