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a_sh-v [17]
3 years ago
6

A person kicks a 4.0-kilogram door with a 48-newton force causing the door to accelerate at 12 meters per second squared. What i

s the magnitude of the force exerted by the door on the person?
Physics
1 answer:
inna [77]3 years ago
7 0

Answer:

-48 N

Explanation:

mass of door (m) = 4 kg

acceleration of the door = 12 m/s^{2}

force exerted by the person = 48 N

From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N

You might be interested in
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2
Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

4 0
3 years ago
Which option identifies the best technique to employ in the following scenario?
ollegr [7]
Answer: Urban Green House
8 0
2 years ago
A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric
sp2606 [1]

Answer:

a)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)E=0

c)E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a)    r < a

At a radius r ,from gauss theorem

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

b)   a < r < b

E.ds=\dfrac{q_i}{\varepsilon _o}

The total induce in this surface = - Q+ Q =0

E.ds=\dfrac{0}{\varepsilon _o}

E = 0

c)   r > b

E.ds=\dfrac{q_i}{\varepsilon _o}

E\times 4\pi r^2=\dfrac{Q}{\varepsilon _o}

E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C

4 0
3 years ago
If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted
Lady bird [3.3K]

Answer:

\frac{Q}{Q_0}=1

Explanation:

Capacitance is defined as the charge divided in voltage.

C=\frac{Q}{V}(1)

Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:

V=\frac{V_0}{k}

Where k is the dielectric constant and V_0 the voltage of the capacitor without a dielectric

The capacitance with a dielectric between the capacitor plates is given by:

C=kC_0

Where k is the dielectric constant and C_0 the capacitance of the capacitor without a dielectric. So, we have:

Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1

Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.

7 0
3 years ago
Help??????????????????????????
hammer [34]

Answer:

imma go with either b or A

3 0
3 years ago
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