Select the correct answer.If the coefficient of static friction is 0.35 and the normal force is 80 newtons, what is the maximum frictional force of the surface acting on the object?
O A.
B.
C.
OD.
OE
9.8 newtons
28 newtons
80 newtons
23 newtons
35 newtons

Answers

Answer 1
Answer:

Final answer:

The maximum frictional force acting on the object is 28 newtons.


Explanation:

The maximum frictional force acting on an object can be calculated using the formula:

Frictional force = coefficient of static friction x normal force

Given that the coefficient of static friction is 0.35 and the normal force is 80 newtons, we can calculate the maximum frictional force as follows:

Maximum frictional force = 0.35 x 80 = 28 newtons


Learn more about Calculating maximum frictional force here:

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A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф=((wi+wf)/(2) )t

Ф=((0+2)/(2) )5=5 rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

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Further explanation

Centripetal Acceleration can be formulated as follows:

\large {\boxed {a = \frac{ v^2 } { R } }

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

\texttt{ }

Centripetal Force can be formulated as follows:

\large {\boxed {F = m \frac{ v^2 } { R } }

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

\texttt{ }

Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

d = \theta R

d = (\omega + \omega_o)(1)/(2)t R

d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )

d = 4\pi (1.75)

d = 7\pi \texttt{ m}

d \approx 22 \texttt{ m}

\texttt{ }

Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.

Answers

Answer:

The force constant of the spring is 735 N/m.    

Explanation:

It is given that,

Mass of fruit, m = 1500 g = 1.5 kg

Compression in the scale, x = 0.02 m

We need to find the force constant of the spring on the scale. The force acting on the scale is given by using Hooke's law. So,

F=-kx

Also, F = mg

mg=kx

k is force constant

k=(mg)/(x)\n\nk=(1.5* 9.8)/(0.02)\n\nk=735\ N/m

So, the force constant of the spring is 735 N/m.

The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁  = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_(1) +q_(2) =-q_(3) -----eqution 1

where,

q_(1) is the heat absorbed by the solid at 0⁰C

q_(2) is the heat absorbed by the liquid at 0⁰C

q_(3) the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_(2) = -q_(3)

Step 4: calculate the final temperature of the water

79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)

Substitute in the values; we will have,

79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})

79.13 kJ + 990.66J* (T_(f)-218}) = -1463J*(T_(f)-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_(f)-218}) = -1.463KJ*(T_(f)-100})

79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3

collect like terms,

2.45366T_(f) = 283.133

T_(f) = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

Which of the following statements correctly compares the relationship between the earth, its atmosphere and radiation?1. The earth is cooled and its atmosphere is heated by solar radiation.

2. The earth is heated and its atmosphere is cooled by terrestrial radiation.

3. The earth is cooled and its atmosphere is heated by terrestrial radiation.

4. The earth is heated and its atmosphere is cooled by solar radiation.

Don't answer unless you know for sure. Thank you so much!

Answers

Answer: The option 4 is correct answer.

Explanation:

Terrestrial radiation is a long wave electromagnetic radiation. It originates from the earth and its atmosphere.

The sun emits a huge amount of energy. It travels across the space. The atmosphere is not directly heated by the solar radiation. It is heated by the terrestrial radiation that the planet itself emits.

When the land is heated then it emits radiation which heats up the atmosphere.

The earth is cooled and its atmosphere is heated by terrestrial radiation.

Therefore, the relationship between the earth, its atmosphere and radiation is correctly compared by statement 4.

The earth is cooled and its atmosphere is heated by terrestrial radiation.