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grigory [225]
3 years ago
14

Help me with this please

Physics
2 answers:
stepan [7]3 years ago
5 0

Answer:

In option D i.e. in 1, 2 and 3

Explanation:

Gravity acts on all objects trying to pull it to the center. It is gravity which keeps us to the ground so gravity basically acts on all the three stages or positions.

Hope you are satisfied with my answer. Hope its helpful for you.

Jobisdone [24]3 years ago
4 0

Answer:

Option D- 1,2 and 3

Explanation:

Gravity always acts on an object irrespective of height

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An object with a momentum of 1500 kg-m/s directed east is acted upon by an impulse of 100.0 kg-m/s in the same direction. What i
fiasKO [112]

Answer:

The final momentum of the body = 1600 kgm/s

Thus the impulse that acted on the body. = 500 N.s

Explanation:

Momentum: This can be defined as the product of mass and velocity. The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Mathematically, momentum can be expressed as

M =mv.

Where M = momentum of the object, m = mass of the object, v = velocity of the object.

Impulse acting on the object = Final momentum of the object - initial momentum of the object

I = M₂ - M₁

M₂ = I + M₁......................... Equation 1

Where I = impulse, M₁ and M₂ = Final and initial momentum.

Note:

(i) The momentum and impulse act in the same direction

(ii) impulse is also a vector quantity.

Given: M₁= 1500 kgm/s, I = 100 kgm/s.

Substituting these values into equation 1

M₂ = 1500 + 100

M₂ = 1600 kgm/s.

Thus the final momentum of the body = 1600 kgm/s

14.

I = m(v-u)............................................... Equation 2.

Where I = impulse on the object, m = mass of the object, v = final velocity, u = initial velocity.

Given: m = 100 kg, v = 15 m/s, u = 10 m/s.

I = 100(15-10)

I = 100(5)

I = 500 N.s or 500 kgm/s.

Thus the impulse that acted on the body. = 500 N.s

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The products of one turn of dark reaction cycle are called
harina [27]
Light Independent Reactions
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A circular loop of wire 78 mm in radius carries a current of 114 A. Find the (a) magnetic field strength and (b) energy density
Finger [1]

Explanation:

It is given that,

Radius of loop, r = 78 mm = 0.078 m

Current, I = 114 A

(a) Magnetic field strength of the circular loop is given by :

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B=\dfrac{4\pi \times 10^{-7}\times 114}{2\times 0.078}

B = 0.000918 T

or

B=9.18\times 10^{-4}\ T

(b) Energy density at the center of the loop is given by :

U=\dfrac{B^2}{2\mu_o}

U=\dfrac{(0.000918)^2}{2\times 4\pi \times 10^{-7}}

U=0.335\ J/m^3

Hence, this is the required solution.

7 0
3 years ago
Please help me with this​
AfilCa [17]

Answer:

20 N exerts no torque about the pivot.

14 N exerts a counterclockwise torque of 14 * .3 = .42 N-m

6  exerts a clockwise torque of 6 * .7 = .42 N-m

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