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3 years ago

how would the velocity of the book change if the applied force were equal to the sliding friction force

1 answer:

3 0

Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>

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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

2 years ago

I will give brainiest to the first person to tell me who is the Carolina panther QB

Inessa [10]

Answer:

Cam Newton (currently but might change because he has been allowed to trade)

Will Grier

Kyle Allen

Explanation:

7 0

3 years ago

Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c

Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0

3 years ago

A cubic tank holds 1,000.0 kg of water. what are the the dimensions of the tank in meters? explain your reasoning

Scilla [17]

Given: A cubic tank holds 1,000.0 kg of water.

Mass of water in tank (m) = 1000.0 kg

Density of water (d) = 1000.0 kg /m³

Concept: Volume(V) = Mass / Density

Since the tank holds these water in it so the volume of water will be equal to the volume of the tank.

Hence, the volume of the tank = Mass of water / Density of water

or, = 1000.0 kg / 1000.0 kg m⁻³

or, = 1.0 m³

Since tank is cubical in shape. Let its side be 'x'

The volume of tank (x³) = 1.0 m³

or. side of tank (x) = 1.0 m

Hence, the dimensions of the tank will be 1.0 m.

8 0

3 years ago

19. Find the recoil velocity of a 65 kg ice hockey goalie who

Alexxx [7]

Answer:

c. 0.12 m/s

Explanation:

by using momentum formula $v_2=\frac{m_1\times v_1}{m_1+m_2}$

we get

$v_2=\frac{0.15\times 50}{0.15+65}=0.12$

4 0

3 years ago