Which free body diagram is in equilibrium?

3. An object with a mass of 10 kg is moving along a horizontal surface. At a certain point it has 40 J of kinetic energy. If the

damaskus [11]

Answer: 100cm

Explanation:

The force of friction on a surface normal to gravity where µ is the coefficient of friction is

F = µmg

Where

F = the friction force

µ = coefficient of friction

m = mass of the object

g = acceleration due to gravity

Also, the Kinetic Energy of the object, E = Fs, where

E = Kinetic Energy

s = stopping distance. So that,

E = µmgs

40 J = 0.4 * 10 kg * 10 m/s² * s

40 J = 40 kgm/s² * s

s = 40 J / 40 kgm/s²

s = 1 m or 100 cm

7 0

3 years ago

The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton

Brrunno [24]

Answer:

A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C) $\frac{F_e}{F_g}$ = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = $6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }$

F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

F = $k \ \frac{q_1q_2}{r^2}$

let's calculate

F = $9 \ 10^9 \ \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}$

F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

$\frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }$

= 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0

3 years ago

Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth

mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

4 0

3 years ago

Read 2 more answers

After a nucleus undergoes radioactive decay, its new mass number is:

Ivanshal [37]

Radioactive "decay" means particles and stuff shoot OUT of a nucleus.
After that happens, there's less stuff in the nucleus than there was before.
So the new mass number is always less than the original mass number.

7 0

4 years ago

Read 2 more answers

A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a ne

blagie [28]

Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

3 0

3 years ago