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disa [49]
3 years ago
7

How much force is needed to accelerate a 1.5 kg physics book to an acceleration of 6 m/s^2?

Physics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

Force = 8.0 k g m / s

Explanation:

Force = mass x acceleration

Mass = 4.0 k g Acceleration = 2.0 m / s 2

Hence,force = ( 4.0 x 2.0 ) k g m / s 2 = 8.0 k g m / s 2

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1. The following can be inferred from Newton’s second law of motion except:
lora16 [44]

Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)

2.a)acceleration decreases

Explanation:

Newton's second law:

Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.

When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.

This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .

7 0
3 years ago
What is the scientific revolution??
Basile [38]
The scientific revolution is a concept which explains how the developments of science (biology, chemistry, physics and etc.) changed the way we (society) think about nature. I hope this helps! :)
4 0
4 years ago
Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?
Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

\frac{Q}{T}  =  \frac{mLf}{T}

Substituting the corresponding values we have:

793.0 kW=  \frac{2.0kg(334 kJ/kg)}{T} \\  T  =  \frac{2.0kg(334 kJ/kg)}{793.0kW}  =  \frac{668kJ}{793kW}   \\  = 0.84s

8 0
3 years ago
A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of
Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0
3 years ago
A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co
nignag [31]

Answer:

Explanation:

Given

mass of box m=80 kg

coefficient of kinetic friction \mu _k=0.15

coefficient of Static friction \mu _s=0.30

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

f_r=ma

where f_r=frictional\ Force

a=acceleration

a=0

f_r=0

4 0
3 years ago
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