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3 years ago

What is the strength of an electric field that will balance the weight of a proton?

Physics

1 answer:

alukav5142 [94]3 years ago

3 0

The weight (force of gravity) on an object near the surface of the earth is given by

F = mg

where m is the object's mass and g is the acceleration of objects near the surface of the earth.

The electric force on a charged object is given by

F = Eq

where E is the electric field at the point where the object is located and q is the charge of the object.

The weight (force of gravity) of a proton will act downwards (toward the surface of the earth), so in order for this force to be balanced, there must be an electric field pointing upward such that the electric force is equal and opposite to the weight. Set the magnitudes of the weight and electric force equal to each other:

mg = Eq

E = mg/q

Known values:

m = 1.67×10⁻27kg (a quick Google search will yield the mass of a proton)

g = 9.81m/s²

q = 1.60×10⁻19C

Plug in the known values and solve for E:

E = 1.67×10⁻27×9.81/(1.6×10⁻19)

E = 1.02×10⁻⁷N/C

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An open system starts with 52 J of mechanical energy. The energy changes

spayn [35]

Answer:

I think the answer is D,54 joules

5 0

3 years ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0

3 years ago

If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final<br>temperature?

Debora [2.8K]

Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

$Q=mc\Delta T$

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

$6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K$

So, the final temperature is equal to 305.17 K.

3 0

3 years ago

During a very quick stop, a car decelerates at 28.4 rad/s?. Assume the tires initially rotated in the positive direction and rad

Damm [24]

Answer:

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

$n = \frac{\theta }{2\pi }$

$n = \frac{152.3 }{2(3.14) }$

$n = 24$

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

8 0

4 years ago

In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres o

SVETLANKA909090 [29]

Answer:

$F = 7.7*10^{-10}N$

Explanation:

You need to be careful with units for this problem. The force will be:

$F =\frac{K*m1*m2}{d^2}$

$F=\frac{6.67259 * 10^{-11}*1.56*21.1*10^{-3}}{(5.34*10^{-2})^2}$

$F=7.7*10^{-10}N$

8 0

3 years ago