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xxTIMURxx [149]
4 years ago
9

A _________________ is when the courts allow people who have all been harmed to pool grievances and sue for damages on behalf of

the group.
a. small claims court
c. class action law suit
b. binding arbitration
d. federal trial
Physics
2 answers:
Vika [28.1K]4 years ago
7 0
C.class action law suit

Sliva [168]4 years ago
4 0

A class action law suit is one in which the courts allow several
people who all claim to have been harmed to pool grievances
and sue for damages on behalf of the group.

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Bicycle in its writers have combined a mass of 80 kg in a speed of 6.0 meters per Second what is the magnitude of the average fo
erik [133]

Answer:

F = 120 N

Explanation:

Force x distance = energy

The bike has energy 1/2 . 80 . 6^2   = 1440 J

You are looking at an example of not reading the question properly.

Impulse = Force . time  = change in momentum

F . 4  = 80 .6

F = 120 N

7 0
3 years ago
Sam is pulling a box up to the second story of his apartment via a string. The box weighs 53.3 kg and starts from rest on the gr
castortr0y [4]

Answer:

W = 1222.4 J = 1.22 KJ

Explanation:

The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:

W = F d Cosθ

where,

W = Work Done = ?

F = Force Applied = 64 N

d = Distance Covered by Box = 19.1 m

θ = Angle between force and displacement = 0°

Therefore,

W = (64 N)(19.1 m)Cos 0°

<u>W = 1222.4 J = 1.22 KJ</u>

7 0
3 years ago
A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

5 0
3 years ago
How will the motion of the arrow change after it leaves the bow?
Pavel [41]

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

4 0
3 years ago
The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35
azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

g= \frac{GM}{R^2}

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

g_{ss} = \frac{GM}{R+h^2}

g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}

g_{ss} = 8.82m/s^2

Therefore the gravitational acceleration at 350km is 8.82m/s^2

5 0
4 years ago
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