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shepuryov [24]
3 years ago
7

Question 8 of 25

Physics
1 answer:
GuDViN [60]3 years ago
7 0

Answer:

A. They use X-rays that penetrate most objects.

You might be interested in
How much net force is required to accelerate a 2000 kg car at 3.00 m/s^2
andrezito [222]

The net force required to accelerate a car is 6000 N.

Force is defined as the product of the mass and acceleration of the body. Force is used to changing the velocity that is to accelerate an object or a body of a particular mass. The unit of Force is Newton or kg m/s^2.

The formula used to calculate the net force is :

F = ma

where, F = Force

m = mass = 2000 kg

a = acceleration = 3.00 m/s^2

∴ F = 2000*3

F = 6000 N

Thus, to accelerate the car at 3.00 m/s^2 of mass 2000 kg net force required is 6000 N.

To learn more about force,

brainly.com/question/1046166

6 0
1 year ago
For the potential , determine the number of bound states?A)6 b)4 c) 3 d) 1
s2008m [1.1K]

Do you see that blank, open space after the word "potential ..." ?

There's supposed to be a number there that actually tells us the value of the potential.  Without that number ... and a lot more description of the whole scenario here ... there's no possible answer to the question.

7 0
3 years ago
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A


Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0
3 years ago
A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The
almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

             Initial momentum = Final momentum

                                       Pi = Pf

                  M*vs_e = m*vc_e + 4m*vm_e

Where,

                  M = m + 4m = 5m

                  vc_e = Velocity of command relative to earth

                  vm_e = Velocity of motor relative to earth  

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

                  vm_e =  v_c/m + vc_e        

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

                  5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

                  5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

                  5m*vs_e -  4m*(v_c/m) = 5m*vc_e

                   vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

                  vc_e = 3760 - 0.8*(90)

                  vc_e = 3,688 km/h

5 0
3 years ago
Read 2 more answers
A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
2 years ago
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