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Ghella [55]
3 years ago
14

A 200-kilogram wood crate is pushed across a wood floor How much force will be needed to move the box at a constant velocity of

0.5 m/s
Physics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

100N

Explanation:

F = m x a

m = 200kg

a = 0.5m/s

F = 200 x 0.5

F = 100

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A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial spe
motikmotik

Answer:

K = G Mm / 9R

Explanation:

Expression for escape velocity V_e = \sqrt{\frac{2GM}{R} }

Kinetic energy at the surface = 1/2 m V_e ²

= 1/2 x m x 2GM/R

GMm/R

Potential energy at the surface

= - GMm/R

Total energy = 0

At height 9R ( 8R from the surface )

potential energy

= - G Mm / 9R

Kinetic energy = K

Total energy will be zero according to law of conservation of mechanical energy

so

K  - G Mm / 9R = 0

K = G Mm / 9R

6 0
3 years ago
Calculate the acceleration of a train travelling from rest to 24 m/s in 12 seconds.
Vilka [71]
Answer: A 2m/s^2

Steps: Formula for acceleration. (Velocity Final - Initial Velocity) / Time

(24 - 0) / 12 = 2
3 0
2 years ago
An extension cord is used with an electric weed trimmer that has a resistance of 17.9 Ω. The extension cord is made of copper (r
Naddika [18.5K]

Answer:

(a) R_{c}=0.87ohms

(b) V_{T}=114.44V

Explanation:

Part (a)

The total length of copper cord L=86.3 m

The cross sectional area A=1.71×10⁻⁶m²

The resistivity of copper p=1.72×10⁻⁸Ω

Thus the resistance of extension cord is

R_{c}=p\frac{L}{A}\\R_{c}=(1.72*10^{-8} )\frac{86.3}{1.71*10^{-6}}\\R_{c}=0.87ohms

Part (b)

The resistance of trimmer Rt=17.9 ohms

When voltage of 120V is applied then the current I is passing through series circuit is

I=\frac{120V}{R_{c} +R_{T} }\\I=\frac{120V}{0.87 +17.9 } \\I=6.4A

Thus the voltage across the trimmer is:

V_{T}=IR_{T}\\V_{T}=(6.4)*(17.9)\\V_{T}=114.44V

8 0
3 years ago
How fast would a 2-kilogram object need to move to have the same kinetic
siniylev [52]

Answer:

11.3 m/s

Explanation:

KE₁ = KE₂

½m₁v₁² = ½m₂v₂²

½ (2 kg) v² = ½ (4 kg) (8 m/s)²

v ≈ 11.3 m/s

8 0
3 years ago
A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

Learn more about work and kinetic energy:

brainly.com/question/6763771  

brainly.com/question/6443626  

brainly.com/question/6536722

#LearnwithBrainly

6 0
3 years ago
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