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Ghella [55]
3 years ago
9

A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067

g d) 0.015 g
Physics
1 answer:
Vesna [10]3 years ago
4 0

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

P = mv

m = \dfrac{P}{v}

Where, P = momentum

v = speed

Put the value into the formula

m = \dfrac{2.8}{187}

m = 0.015\ kg

m = 15\ g

Hence, The mass of the bullet is 15 g.

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Its letter C. 5N to the left. Since Jeremy's force in Newtons are higher than Amanda's (in newtons), and since Jeremy's force directs to the left, then the direction of the force will be to the LEFT. Then subtract the higher one to the lower one so that would be: 10N-5N=5N. So it is C. 5N to the left.
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1pt If the room is the frame of reference for the study of an object moving in three dimensions, from where should distances be
oksian1 [2.3K]

Answer:

Option B, two walls and the floor

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The distance should be measured from the point where at least the three axed meet.

Two walls and the floor are equivalent to three axes.

Vertical wall 1 = Y Axis

Horizontal wall2 = X Axis

Floor = Z Axis

Thus, the distance should be measured from the point where two walls and one floor meet.

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What will happen if position of an effort and load are interchanged in the hydraulic press?why?
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Will load input interchange

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A parallel-plate air capacitor of area A = 15.0 cm2 and plate separation d = 3.00 mm is charged by a battery to a voltage 58.0 V
sladkih [1.3K]

Answer:

The additional charge that will flow from the battery onto the positive plate is 0.924 nC

Explanation:

Given;

Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²

Separation distance, d = 3.00 mm = 3 x 10⁻³ m²

voltage of the capacitor, V = 58.0 V

dielectric constant, k = 4.60

Initial Capacitance of the capacitor before the addition of dielectric material:

C = \frac{\epsilon _oA}{d} = \frac{8.85*10^{-12}*15*10^{-4}}{3*10^{-3}} = 4.425 *10^{-12} F

Initial charge across the parallel plates:

Q₁ = CV

Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C

Capacitance of the capacitor after the addition of dielectric material:

Cequ. = Ck

Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F

Final charge across the parallel plates:

Q₂ = Cequ. x V

Q₂ =  20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C

Additional charge = Q₂ - Q₁

                              = 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C

                              = 9.239 x 10⁻¹⁰ C

                              = 0.924 nC

Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC

3 0
3 years ago
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