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4 years ago

Haloke does 176 J of work lifting himself 0.300m at a constant speed. What is Haloke's mass?

1 answer:

Colt1911 [192]4 years ago

5 0

Since he's lifting himself at constant speed ... not accelerating ... the net vertical force on
him is zero, meaning that he must be lifting with force that's exactly equal to his weight.

Potential energy he added during the lift = (mass) x (gravity) x (height)

Mass = (energy) / (gravity x height) = 176 / (9.8 x 0.3) = 59.864 kg

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Help me plzzz I need answers

soldier1979 [14.2K]

Answer:

i think it is B

Explanation:

4 0

3 years ago

A crate is being pulled down an incline as shown in the figure. With respect to the crate's direction of motion, which of the fo

Dmitrij [34]

Answer: Fn

Explanation: Because Fn is applying force upward

7 0

3 years ago

A couch of mass 210 kg is being pushed across the floor with an applied force of 4100 N. The coefficient of kinetic friction bet

Dennis_Churaev [7]

Let us use the formula for Newton's Second Law of Motion:

Net force = Mass*Acceleration
Net force = Applied Force - μ*Normal Force
where μ is the coefficient of kinetic friction

Normal Force = Force due to gravity = mass*gravity
Normal Force = (210 kg)(9.81 m/s²) = 2,060.1 N

Then,
Net force = 4100 - 0.38*2060.1 = 3317.162 N

3317.162 N = (210 kg)(a)
Solving for acceleration,
a = 15.796 m/s²

8 0

4 years ago

A 17 kg box experiences an applied force of +175 N and a force of friction of -125 N. While experiencing these unbalanced forces

Flura [38]

Answer: (b)

Explanation:

Given

Mass of box $m=17\ kg$

Applied force $F=175\ N$

Friction force $f=-125\ N$

Box travels a distance of $s=15\ m$

time taken $t=5.25\ s$

Net unbalanced force $F_{net}=175-125=50\ N$

Work done by the unbalanced force

$\Rightarrow W=F_{un}\cdot s\\\Rightarrow W=50\times 15\\\Rightarrow W=750\ N$

Power developed by unbalanced force

$\Rightarrow P=\dfrac{W}{t}\\\\\Rightarrow P=\dfrac{750}{5.25}\\\\\Rightarrow P=142.85\approx 1423\ W$

Thus, option (b) is correct.

7 0

3 years ago

A car of mass 650 kg slows from a speed of 20.0 m/s to a speed of 10.0 m/s. How much kinetic energy has it lost. How much force

Natalija [7]

Formulas you need for this problem:
F= mass•acceleration
KE= mass•velocity^/2
Acceleration= final velocity-intial velocity/time
time= distance/speed

t= 97.5\10= 9.75 seconds
Acc= 10-20/9.75= -1.03 m/s/s

These are the answers :)
F=650•-1.03= -669.5N
KE= 650•100/2= 32,500J or 32.5KJ

7 0

4 years ago