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A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t

MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity $u_x$ . It takes a time t to fall a vertical distance y and at the same time travels a horizontal distance x.

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance y using the equation,

$y=u_yt+\frac{1}{2} gt^2$

Substitute 0 m/s for $u_y$ , 9.81 m/s²for g and 1.5 m for y.

$y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s$

The horizontal distance traveled by the bullet is given by,

$x=u_xt$

Substitute 500 m/s for $u_x$ and 0.5530s for t.

$x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m$

The bullet travels a distance of 276.5 m.

5 0

3 years ago

An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De

gayaneshka [121]

We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
$F=T-W$
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
$a= \frac{F}{m}$
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
$F=T-W=6000 N-5000 N=1000 N$
To find m, we can use the weight of the system. In fact, the weight of an object is given by
$W=mg$
where $g=9.81 m/s^2$ . Solving for m, and using W=5000 N, we find
$m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg$

and at this point, we can calculate the acceleration of the system (elevator+people):
$a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2$
and the acceleration has the same direction of the resultant force, so upward.

8 0

3 years ago

A tennis player hits a ball 2.0 m above the ground.

tangare [24]

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

As it hits the ground in time t, so put y = 0

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

your welcome, and have a great day.

8 0

3 years ago

Read 2 more answers

Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)

Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

$\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day$

Us the equation of radioactivity

$N=N_{0}e^{-\lambda t}$

$0.1N_{0}=N_{0}e^{-0.08664 t}$

$e^{0.08664 t}=10$

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0

3 years ago

After being accelerated through a potential difference of 5.33 kv, a singly charged carbon ion (12c) moves in a circle of radius

LekaFEV [45]

I'm stuck on the same question, as well :(

4 0

3 years ago