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zaharov [31]
3 years ago
13

OThy look jump mo I’ll I’ll

Physics
1 answer:
madreJ [45]3 years ago
6 0
I have no idea what this means but I hope you are having a lovely day and that life treats you wonderfully because everyone deserves happiness. I hope you reach your goals that you’re pursuing. Have a nice day :)
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SP: Calculate the moment
ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, F=80\ N

Distance of application of force from the point about which moment is needed is, d=25\ cm=\frac{25}{100}\ m=0.25\ m

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0
3 years ago
4 WILL GIVE YOU BRAINLIST
Arisa [49]
The last one, handmade gifts require more of the givers time!
5 0
3 years ago
Read 2 more answers
A car is moving in a straight line with the same speed of 100 m/s ,the acceleration in this
Amiraneli [1.4K]

Answer:

a = 0m/s²

Explanation:

Average acceleration = (change in velocity)/(time it takes). Since the car's change in velocity is zero, its acceleration is zero.

4 0
3 years ago
A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop
postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination(\theta )=30^{\circ}

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

F_{net}=0

Force applied by cable=mg\sin \theta

F=60\times 9.8\times \sin (30)=294 N

work done=F\cdot x

W=294\cdot 75=22.125 J

(b)Power=F\cdot v

P=294\cdot 2.4=705.6 W\approx 0.946 hp

5 0
3 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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