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LiRa [457]
3 years ago
12

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

?
Physics
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 spaceship, at rest in a certain reference frame s, is given a speed increment of 0.500c. it is then given a further 0.500c inc
bezimeni [28]

In order to give a spaceship at rest in a specific reference frame s a speed increment of 0.500c, seven increments are required. Then, in this new frame, it receives an additional 0.500c increment.

The speed of an object, also known as v in kinematics, is a scalar quantity that refers to the size of the change in that object's position over time or the size of the change in that object's position per unit of time. The distance travelled by an object in a certain period of time divided by the length of the period gives the object's average speed in that period.

The spacecraft moves at v1 = 0.5c after the initial increment.The equation becomes V2 = V+V1/1+V*V1/c after the second one. 2 V2 = 0.5c+0.50c/1+(0.50c)^2/c^ 2 = 0.80c

Likewise, V3 = 0.929c

V4 = 0.976c

V5 = 0.992c

V6 = 0.99c

V7 = 0.999c

Learn more about speed here

brainly.com/question/28224010

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5 0
1 year ago
Two wires each carry 10.0 A of current (in opposite directions) and are 2.50 mm apart. What is the magnetic field 37.0 cm away a
lyudmila [28]

Answer:

see answer below

Explanation:

Before we do any kind of calculation, we need to convert the proper units of the exercise. All the units of distance must be in meters, so, let's change distance of the wire, and the magnetic field to meters:

Separation between the wires are 2.5 mm:

2.5 mm * (1 m / 1000 mm) = 0.0025 m

The distance of P from the bottom of the wires is 37 cm:

37 cm * (1 m/100 cm) = 0.37 m

The distance of P from the top of the wires is just the sum of the two distances:

R = 0.37 + 0.0025 = 0.3725 m

Now that we have the distance, we can determine the magnetic field, using the following expression:

B = B(bottom) - B(top)   or just B₂ - B₁

And B = μ₀ I / 2πR

Replacing in the above expression we have:

B = μ₀ I / 2π ( 1/R₂ - 1/R₁)

Now we can determine the magnetic field:

B = (4πx10⁻⁷ * 10 / 2π) (1/0.37 - 1/0.3725)

<h2>B = 3.63x10⁻⁸ T</h2><h2></h2>

Which means that the magnetic field is out of the page.

Hope this helps

4 0
3 years ago
Which of the following does not illustrate a complete circuit?
Yakvenalex [24]

Answer:

I think A

Explanation:

because it dosn't have enough tools

4 0
2 years ago
Calculate the potential energy of a +1.0μC point charge sitting 0.1m from a -5.0μC point charge.
AfilCa [17]

Answer:

P.E.      =   -0.449 J

Explanation:

Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.

Now Potential energy is defined by this formula,

P.E. = k q₁ q₂/ r

where P.E. is the potential energy.

k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)

q₁ = charge of one particle = +1.0μC

q₂ = charge of another particle = -5.0μC

r = distance = 0.1 m

Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m

          P.E.      =  -0.449 J

8 0
3 years ago
:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?​
denis-greek [22]
Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both
3 0
3 years ago
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