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Romashka-Z-Leto [24]
3 years ago
10

Your physics teacher is giving you a special gift a box of mass 20 kg with a mystery surpise insie .the box is placed on a frict

ionsed table .what is the weigth of the box and normal reaction over it by the table ?
Physics
2 answers:
TiliK225 [7]3 years ago
7 0

Answer:

F = 196.2 N

Explanation:

Use Newtons law

F = m * a

F = 20 kg * 9.81 m/s²

F = 196.2 N

Debora [2.8K]3 years ago
4 0

Answer:

\Huge \boxed{\mathrm{196.2 \ N}}

\rule[225]{225}{2}

Explanation:

Using weight formula:

\sf Weight = mass \times acceleration \ of \ gravity

W=mg

The mass of the box is 20 kg.

The acceleration of gravity on Earth is 9.81 m/s².

W=(20)(9.81)

W=196.2

The weight of the box is 196.2 N.

The net force of the box is zero, since the box is at rest. The normal force is exerted upward on it by the table.

\rule[225]{225}{2}

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True or false when two containers at the same temperature are brought together, no heat is transferred
Murrr4er [49]
<span>when two containers at the same temperature are brought together, no heat is transferred.

It is true
</span>
6 0
3 years ago
A meteoroid is moving towards a planet. It has mass m= 0.18x10⁹ kg and speed v1 = 5.9x10⁷ m/s at distance R1 = 2.9x10⁷ m from th
anzhelika [568]

Answer:

5.90000114e+7 m/s

Explanation:

Given values

M₁ = 1.8e+8 kg

M₂ = 2.4e+25 kg

M = M₁ + M₂ = 2.400000000000000018e+25 kg

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

GM = 1.60183200000000001201374e+15 m³ sec⁻²

r₁ = 2.9e+7 m

v₁ = 5.9e+7 m/s

R = 2.2e+6 m

The total energy of the meteroid per unit mass at time t₁ is

E/M₁ = 0.5v₁² − GM/r₁ = 1.7404999447644137931034478615952e+15 m²/s²

At impact,

E/M₁ = 0.5v₂² − GM/R = 1.7404999447644137931034478615952e+15 m²/s²

v₂ = √[2(E/M₁ + GM/R)] = 5.9000011404572937396882314817886e+7 m/s

v₂ − v₁ ≈ 11.4 m/s

4 0
3 years ago
Rank the six combinations of electric charges on the basis of the electric force acting on q1.
loris [4]

Answer:

Largest; options A & B

2nd largest; Option F

3rd Largest; Option D

Smallest; Options E & C

Explanation:

Looking at the charges, basically, the net force of the charge q1 is the sum of charges q2 and q3.

When 2 charges are opposite, the force will be attractive but if they are same, then the force will be repulsive.

Thus, the order of the forces is;

I) Largest: q1= +1nC, q2= -1nc, q3= -1nc and (q1= -1nC, q2= +1nc, q3= +1nc)

II) Second largest is: q1= +1nC, q2= -1nc, q3= +1nc

III) Third largest: q1= +1nC, q2= +1nc, q3= -1nc

IV) Smallest: (q1= -1nC, q2= -1nc, q3= -1nc) and (q1= +1nC, q2= +1nc, q3= +1nc)

6 0
3 years ago
find the gravitational force this shell exerts on a 1.60 kg point mass placed at the distance 5.01 m from the center of the shel
RideAnS [48]

The gravitational force of the shell exerts is 4.25m x 10¯¹² N.

We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses of objects. The magnitude of gravitational force can be determined as

F = G.m1.m2 / R²

where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.

From the question above, we know that

m1 = 1.6 kg

m2 = m

R = 5.01 m

By substituting the following parameters, we get

F = G.m1.m2 / R²

F = 6.674 × 10¯¹¹  . 1.6 . m / 5.01²

F = 4.25m x 10¯¹² N

where m is the mass of the shell

For more on gravitational force at: brainly.com/question/19050897

#SPJ4

7 0
1 year ago
Calculate δe, if the system absorbs 7.24 kj of heat from the surroundings while its volume remains constant (assume that only p−
butalik [34]
I believe the answer is 7.24 kJ.
From the equation ΔE = dW + dQ; where W is the work done on/by the system and Q is the heat the system absorbs/loses.
Therefore; ΔE = 72.4 kJ since the system has bot done any p-v work (dV= zero) and has absorbed heat.
4 0
3 years ago
Read 2 more answers
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