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Nataly [62]
3 years ago
10

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons

tant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s
Physics
1 answer:
torisob [31]3 years ago
6 0

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

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According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this
Andrew [12]

Answer:

Part a)

\rho = 0.55 g/cm^3

Part b)

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

F_b = \rho_w V_{sub} g

F_b = (1 g/cm^3) (30 - 13.5) Ag

F_b = 16.5 Ag

Now we know that the weight of the cylinder is given as

W = \rho (30 cm)A g

now we have

\rho (30 cm) A g = 16.5 A g

\rho = 0.55 g/cm^3

Part b)

When the same cylinder is floating in other liquid then we will have

F_b = \rho_L (30 - 18.9 )A g

so we have

\rho_L (11.1) Ag = 0.55(30) Ag

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0
3 years ago
A car is behind a truck going 25 m/s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car ca
user100 [1]

Answer:

No he should not attempt the pass

Explanation:

Let t be the time it takes for the car to pass the truck. The driver should ONLY attempt to pass when the distance covered by himself plus the distance covered by the oncoming car is less than or equal 400 m (a near miss)

At acceleration of 1m/s2 and a clear distance of 10 + 20 + 10 = 40 m, we can use the following equation of motion to estimate the time t in seconds

s = at^2/2

40 = 1t^2/2

t^2 = 80

t = \sqrt{80} = 8.94 s

Within this time frame, the first car would have traveled a total distance of the clear distance (40m) plus the distance run by the truck, which is

8.94 * 25 = 223.6m

So the total distance traveled by the first car is 223.6 + 40 = 263.6m

The distance traveled by the 2nd car within 8.94 s at rate of 25m/s is

8.94 * 25 = 223.6 m

So the total distance covered by both cars within this time frame

223.6 + 263.6 = 487.2m > 400 m

So no, he should not attempt the pass as we will not clear it in time.

8 0
3 years ago
How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?
ioda

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

3 0
3 years ago
Write the mathemetical relation between work force and displacement​
il63 [147K]

Answer:

Work is measured as the product of force and the displacement in the direction of the force. Work = force × displacement in the direction of the force.

8 0
2 years ago
SOMEBODY PLEASE HELP!!! Indicate the reasons why the centripetal acceleration (and centripetal force) always point to the center
maxonik [38]
Well we know the correct answer cannot be "a" bcause velocity is tangent to the circlular path of an object experienting centripical motion. Velocity DOES NOT point inward in centripical motion.

we know the correct answer cannot be "b" because "t" stands for "time" which cannot point in any direction. so, time cannot point toward the center of a circle and therefore this answer must be incorrect.

I would choose answer choice "c" because both force and centripical acceleration point toward the center of the circle.

I do not think answer choice "d" can be correct because the velocity of the mass moves tangent to the circle. velocity = (change in position) / time. Therefore, by definition the mass is moving in the direction of the velocity which does not point to the center of the circle.

does this make sense? any questions?

3 0
3 years ago
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