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2 years ago

15

You pull with a force of 255 N on a piece of luggage of mass 68 kg, but it does not move. If the coefficient of friction between

the luggage and the floor is 0.7, what is the force of static friction acting on the luggage?

2 answers:

VLD [36.1K]2 years ago

8 0

Answer:

the other guy was right, the answer is 255

Explanation:

STALIN [3.7K]2 years ago

3 0

The force of static friction acting on the luggage =255 N

Explanation:

Applied force= 255 N

mass of luggage=68 kg

coefficient of friction= μ=0.7

Normal force=N = mg

N= 68(9.8)=666.4 N

now the force of friction is given by

Ff= μN

Ff=0.7 (666.4)

Ff=466.5 N which is greater than the applied force.

When the applied force is less than the force of friction, in that case

force of friction= applied force

Thus the force of station friction acting on the luggage = 255 N

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A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.

juin [17]

A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.0 m/s.
After the collision the raft moves to the left at 14.4 m/s assuming water simulates a frictionless surface.
Mass of the canoe (m1) = 16 Kg
Mass of the raft (m2) = 14 Kg
Initial velocity of the canoe (u1) = 12.5 m/s
Initial velocity of the raft (u1) = - 16 m/s [Here, the raft's velocity is negative, because the objects are moving in the opposite direction]
Total momentum of the system = m1u1 + m2u2 = [(16 × 12.5) + (14 × -16)] Kg m/s = (200 - 224) Kg m/s = -24 Kg m/s
Final velocity of the raft (v2) = 14.4 m/s
Let the final velocity of the canoe be v1.
Total momentum of the system after the impact = m1v1 + m2v2 = [(16 × v1) + (14 × 14.4)] Kg m/s = 16v1 Kg + 201.6 Kg m/s
According to the law of conservation of momentum, Total momentum of the system before the impact = Total momentum of the system after the impact
or, -24 Kg m/s = 16v1 Kg + 201.6 Kg m/s
or, -24 Kg m/s - 201.6 Kg m/s = 16v1 Kg
or, -225.6 Kg m/s = 16v1 Kg
or, v1 = -225.6 Kg m/s ÷ 16 Kg
or, v1 = -14.1 m/s

<u>Answer:</u>

<u>T</u><u>he final velocity of the </u><u>canoe </u><u>is </u><u>-</u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>or </u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>to </u><u>the </u><u>right.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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2 years ago

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

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2 years ago

A parallel-plate capacitor in air has a plate separation of 1.51 cm and a plate area of 25.0 cm2. The plates are charged to a po

alexdok [17]

Answer:

(a) 380.96 pC

(b) 3.25V

Explanation:

(a) Before immersion,

$C_{air}$ = $\frac{E_{0}A }{d}$

⇒(8.85E-12× 25E-4× 260) ÷(0.0151)

= 380.96 pC

(b) Charge on the plates after immersion can be calculated by,

Q = ΔV×C

= Δ $V_{air}$ ÷ K

where K is the constant for distilled water

= 260 ÷ 80

= 3.25V

6 0

2 years ago

Read 2 more answers

Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of ma

Alex777 [14]

Answer:

N₂ = 503.8 N

Explanation:

given,

mass of bottom block = 37 Kg

mass of middle block = 18 Kg

mass of the top block = 16 Kg

force acting on the top block = 170 N

force on the block at top

N₁ be the normal force from block at middle

now,

N₁ = 170 + m g

N₁ = 170 + 16 x 9.8

now, force on block at middle

N₂ be the normal force exerted by the bottom block

N₂ = N₁ + m₂ g

N₂ = 326.8 + 18 x 9.8

N₂ = 503.8 N

hence, normal force by bottom block is equal to N₂ = 503.8 N

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3 years ago

What color will the paper appear? Explain why?

Evgen [1.6K]

Answer:

The answer is B

Explanation:

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2 years ago