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leonid [27]
2 years ago
7

A bullet is fired from a gun at 45° angle to the horizontal with a velocity of 500 m/s. Find the

Physics
1 answer:
rjkz [21]2 years ago
7 0

answer :

D. 6370.92 m

Explanation:

pls refer to the attachment...

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How high can a 40 N force move a load, when 395 J of work is done?
Juliette [100K]

Answer:

9.875

Explanation:

w=f×s

395=40×s

make s the subject of the formula

s=395/40

=9.875

7 0
3 years ago
) The centre of a mass, m, is at a distance, r, from the centre of a larger mass M. Assuming there are no other masses present,
ivanzaharov [21]

Answer:

U = - G m M / r

Explanation:

The gravitational potential energy is given by the expression

         U = - G m₁ m₂ / r

dodne G is the gravitational cosntnate (G = 6.67 10⁻¹¹¹), m and m are the mass of the bodies involved

subtype the given values

         U = - G m M / r

8 0
3 years ago
It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?
ankoles [38]
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed 
 Vf = 27.5 m/s  Final speed 

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2  -24.0 ^2)
160kj =  4680 x m
convert kilo joules to jeoules                     160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
4 0
3 years ago
A 250 kg car has 6875 kg•m/s of momentum. What is it’s velocity?
liraira [26]

Answer:

v = 27.5 m/s

Explanation:

p = m × v

6,875 = 250 × v

250v = 6,875

v = 6,875/250

v = 27.5 m/s

5 0
2 years ago
A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3
shusha [124]

Answer:

a

  x_2 = -2.3356

b

 v = -1.384 \ m/s

Explanation:

From the question we are told that

  The initial position of the particle is  x_1 = 0.180 \ m

  The initial  velocity of the particle is  u = 0.060  \  m/s

  The acceleration is   a = -0.380 \  m/s^2

   The time duration is  t = 3.80  \ s

Generally from kinematic equation

    v = u + at

=>  v = 0.060 + (-0.380 * 3.80)

=>  v = -1.384 \ m/s

Generally from kinematic equation

   v^2 = u^2 + 2as

Here s is the distance covered by the particle, so

   (-1.384)^2 = (0.060)^2 + 2(-0.380)* s

=>  s = -2.5156 \ m

Generally the final position of the particle is  

    x_2 = x_1 + s

=>   x_2 = 0.180 + (-2.5156)

=>   x_2 = -2.3356

6 0
2 years ago
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