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Mumz [18]
3 years ago
12

Suppose you are driving in a convertible with the top down at constant speed. You point a rifle straight upward and fire it. In

the absence of air resistance, would the bullet land
(a) behind you,
(b) ahead of you, or
(c) in the barrel of the rifle?
Physics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

(c) in the barrel of the rifle

Explanation:

When we are moving in a convertible with constant speed so as per inertia all the objects are moving with same speed as that the speed of convertible.

So here we can say that bullet when shot from the rifle then its speed is same as that of convertible in horizontal direction

So here we can say that if air friction is absent then bullet will move same horizontal distance as the distance moved by you in the convertible

so after it it will reach again at the same position then the displacement of you and bullet is same in horizontal direction so we can say that correct answer will be

(c) in the barrel of the rifle

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If a machine will do 50 J of work in 35 seconds how much power did it produce?
grandymaker [24]
Power is the amount of work done over a period of time. If you will put that into an equation, the formula of power will be:

P = W/t
Where:
P = power
W=work
t = time

Your problem already provides you with work and time so all you need to do is divide:

P = W/t
P = 50J/30s
P = 1.67 W
6 0
2 years ago
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
2 years ago
1. A rock of granite has a mass of 50 kg. if it’s weight in water
igor_vitrenko [27]

Answer:

The first part of the question is asking about BUOYANT FORCE or UPTHRUST.

Upthrust =TRUE WEIGHT-APPARENT WEIGHT

TRUE WEIGHT=mg

TRUE weight=50kg×10m/s²

=500N

upthrust=500N-380N

FB=120N

volume of the rock=mass/density.

since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.

upthrust=Vdg

120N=V×1000kg/m³×10m/s²

120N=V×10000kg/m²s²

120/10000=V

v=0.012m³

please mark brainliest, hope it helped

6 0
2 years ago
A 12.65 g sample of a radioactive substance is allowed to decay for 17.22 min. At that time, the sample weighed 3.115 g. What is
Mademuasel [1]

Answer:

8.61 min

Explanation:

original mass= 12.65

first half life = 12.65/2 = 6.325

second half life = 6.325/2 = 3.1625

Note : 3.1625 is the closest to the value (3.115) given so we work with it

total time for decay =17.22

therefore two decays = 17.22/2= 8.61

8 0
3 years ago
A student pulls a 50-newton sled with a force having a magnitude of 15 newtons. What is the magnitude of the force that the sled
Simora [160]

Answer:

Force = 35 N

Explanation:

From Newton's third law of motion, the boy must apply a force greater than the weight of the sled to lift it.

weight of sled = mg

where m is its mass and g the force of gravity on it.

weight of sled = 50 N

Force applied by the boy on the sled = 15 N

Since the force applied on the sled by the boy is lesser than the weight of the sled, then;

Force that the sled exerts on the student = 50 - 15

                                             = 35 N

The force exerted by the sled on the student is 35 N.

5 0
3 years ago
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