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hodyreva [135]
3 years ago
14

How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?

Physics
2 answers:
inn [45]3 years ago
8 0

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Montano1993 [528]3 years ago
4 0

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Explanation:

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iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig
Alika [10]

Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm  in air

8 0
2 years ago
An open cart is rolling to the left on a horizontal surface. A package slides down a chute and lands in the cart. Which quantity
Marat540 [252]

Answer: the horizontal component of total momentum

Explanation:

Since the open cart is rolling to the left on the horizontal surface, the quantity that has the same value just before and just after the package lands in the cart is the horizontal component of total momentum.

Momentum, is the product of the mass of a particle and the velocity of the particle. The change of momentum depends on the force which acts on it. The addition of the the individual momenta is the total momentum.

8 0
3 years ago
A 10 kg box is 1.3 m above the ground. How much potential energy does it have? (g on Earth of 9.8 m/s?
Volgvan
Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J
8 0
3 years ago
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le
kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0
3 years ago
Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of
kolbaska11 [484]

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

7 0
3 years ago
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