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neonofarm [45]
3 years ago
7

Which statement is correct about the car?

Physics
2 answers:
JulijaS [17]3 years ago
6 0

Answer:

B

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

olga55 [171]3 years ago
5 0

Answer:

The car is moving forward

Explanation:

=> Forward Force = Force on the engine = 2000 N

=> Backward Force = Air Resistance + Friction = 800+500 = 1300 N

Since the forward force is more, the car moves in the forward directing fighting with the air resistance as well as friction.

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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
9. What is the gravitational potential energy of a 61.2 kg person standing on the roof of a 10-storey building? (Each storey is
kow [346]

Answer:

15009

Explanation:

PE = mgh

PE = 61.2(9.81)(10 * 2.50)

PE = 15009.3

3 0
2 years ago
A strong inductive argument must have true premises<br><br> True<br><br> False
Amiraneli [1.4K]
That is true imo not false
6 0
3 years ago
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia
masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

V(t) = \epsilon. e^{-t.\frac{L}{R} }

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

0.8*19 = 19. e^{-4.\frac{22}{R} }

0.8 = e^{-\frac{88}{R} }

ln(0.8) = ln(e^{-\frac{88}{R} })

ln(0.8) = -\frac{88}{R}

R = -\frac{88}{ln(0.8)}

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0
3 years ago
A 2-kg box sits on a horizontal table. the force of friction between the box and the table is 10 n. the box is pushed to the rig
dangina [55]
By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.

Rearranging this equation to find acceleration would give us:
a = F/m

The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N

The mass is 2kg.

Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2

The acceleration of the box is 5ms^-2
6 0
3 years ago
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