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kow [346]
3 years ago
8

Which value is equivalent to 7.2 kilograms?

Physics
2 answers:
ivanzaharov [21]3 years ago
5 0

Explanation:

B.7,200 grams.....................

ra1l [238]3 years ago
5 0

Answer:

B 7200 grams

As

7.2 kg = 7.2 * 1000 gram = 7200 grams

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You pull with a force of 77 N on a piece of luggage of mass 23 kg, but it does
Vinvika [58]

Answer:

The force of static friction acting on the luggage is, Fₓ = 180.32 N

Explanation:

Given data,

The mass of the luggage, m = 23 kg

You pulled the luggage with a force of, F = 77 N

The coefficient of static friction of luggage and floor, μₓ = 0.8

The formula for static frictional force is,

                                      Fₓ = μₓ · η

Where,

                                  η - normal force acting on the luggage 'mg'

Substituting the values in the above equation,

                                   Fₓ = 0.8 x 23 x 9.8

                                        = 180.32 N

Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N

5 0
3 years ago
Read 2 more answers
Two concentric, coplanar, circular loops of wire of different diameter carry currents in the same direction. Describe the nature
Ierofanga [76]

Answer:

Both will be attractive in nature.

Explanation:

In the given case, the direction of the magnetic field is same in both loops as the direction of the current is same in both loops. When two parallel straight wire carrying current in the same direction are brought close to each other, the force between them is attractive in nature. In the same way,  when two coplanar, circular and concentric loops of wire are carrying current in the same direction, the force between them is attractive in nature. It can be checked by using right hand thumb rule.

Check out other explanations.

brainly.com/question/15555539

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4 0
2 years ago
A circuit has a voltage of 10 V and a current of 5 A. What must the resistance be?
azamat

Answer:

R=V/I

R= 2

Explanation:

R = 10V/5A

R = 2ohms

3 0
3 years ago
A current-carrying loop of wire lies flat on a horizontal tabletop. When viewed from above, the current moves around the loop in
Oksana_A [137]

Answer: i dont do physics yet lol

Explanation:

6 0
3 years ago
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

3 0
3 years ago
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