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3 years ago

Parallel to the handle, what is the force of friction acting on the suitcase?

1 answer:

5 0

<span>Lets call F the friction force which will act horizontally backwards.

As you are travelling at a constant velosity horizontally there is no overall resultant force in this direction.

ie. the force you pull with will be equal to the friction force resisting you. (you will initially have to have pulled with a greater force than the friction to get the suitcase moving)

the value of your force pulling is 60 cos26.9 (horizontally) - you should have learnt about resolving forces.

this must be equal to F

so

F=60cos26.9
F=53.5N

hope this helps you
please mark this as brainliest answer</span>

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A crate slides off the back of a flatbed truck that accelerated forward from rest. If the bed of the truck is frictionless, the

jolli1 [7]

The right answer should be B

5 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

Describe Kinetic Energy.

mina [271]

In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

4 0

3 years ago

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

nataly862011 [7]

Answer:

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load $m_{LS}$ = 10000 kg

mass of flat bed $m_{FB}$ = 20000 kg

initial speed of truck $v_{0}$ = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs $F_{N}$ -------------let this be equation 1

where $F_{N}$ = normal force = mg

Fs,max = μs mg

ma $_{max}$ = μs mg

divide through by mass

a $_{max}$ = μs g ---------- let this be equation 2

in equation 2, we substitute in our values

a $_{max}$ = 0.5 × 9.8 m/s²

a $_{max}$ = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

$v_{f}$ ² = $v_{0}$ ² + 2aΔx

where $v_{f}$ is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² / 9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0

3 years ago

A car has a mass of 1,200 kg. What is its acceleration when the engine exerts a force of 600 N? (Formula: F=ma)

Rasek [7]

Answer:

Option (a)

Explanation:

Given that,

Mass of a car, m = 1200 kg

Force exerted by the engine, F = 600 N

Noe force,F = ma

a is the acceleration of the engine

$a=\dfrac{F}{m}\\\\a=\dfrac{600\ N}{1200\ kg}\\\\a=0.5\ m/s^2$

So, the acceleration of the car is 0.5 m/s².

6 0

3 years ago