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viktelen [127]
3 years ago
12

A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo

r the bullet to hit the target ?
Physics
1 answer:
Alex3 years ago
5 0

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

   Speed = \frac{distance}{time taken}

   Since time is unknown, we make it the subject of the expression;

   time = \frac{distance }{speed} = \frac{400}{600}

   time = 0.67s

Learn more:

Speed brainly.com/question/10048445

#learnwithBrainly

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A plane travels down a runway 2750 m before it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1
ss7ja [257]

Answer: 4.236km

Explanation:

Let's define the point (x, y) as:

x = horizontal distance moved.

y = vertical distance moved.

If the plane starts in the point (0, 0) then:

"A plane travels down a runway 2750 m before it lifts off..."

At this time, the position will be:

P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

In this case, as the angle is measured from the horizontal, the components will be:

x = 1.8km*cos(37°) = 1.438km

y = 1.8km*sin(37°) =  1.083 km

Then the new position is:

P = (2750m + 1.438 km, 0 + 1.083 km)

Let's write it using the same units for all the quantities:

we know that

1km = 1000m

Then:

2750m = (2750/1000) km = 2.750 km.

Then we can write the new position as:

P = (2.750 km + 1.438km, 1.083km) = (4.188km, 1.083km)

Now, we define the displacement as the distance between the final position and the initial position.

The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

(0, 0) for the initial position

(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

5 0
2 years ago
If the 50 kg objects slows down to a velocity of 1 m/s how much kinetic energy does it have?
Bogdan [553]

It has 50kg with a velocity of 1 m/s times the speed of the cart divided by 2 and multiplied by kinectic x plus 5

3 0
3 years ago
Read 2 more answers
A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
Leya [2.2K]

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

8 0
3 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
which changes of state should she place in the column titled "lose energy"?freezing, boiling, and meltingmelting, vaporization,
Blababa [14]
Freezing (liquid to solid)
Deposition (gas to solid)
Condensation (gas to liquid)

All three of these state changes are a result of a energy loss. When considering energy loss it is best to think of situations where temperature has dropped. Less energy in the system results in less energy the substance is exposed to or has available. 
7 0
2 years ago
Read 2 more answers
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