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Lera25 [3.4K]
3 years ago
15

In which situation can you be at rest and moving at the same time?

Physics
2 answers:
mojhsa [17]3 years ago
8 0
I think the answer is ( When riding a bike...)
xxTIMURxx [149]3 years ago
7 0
I believe the last one about the bike is correct.
You might be interested in
A father racing his son has 1/2 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.4 m
BaLLatris [955]

Answer: a) 0.78 m/s b) 1.57 m/s

Explanation:

M = father's mass


m = son's mass = M/3


V = father's initial speed


v = son's initial speed

(1/2)MV^2 = (1/2)*(1/2)*m v^2


M*V^2 = (1/2)(M/3)v^2


V^2/v^2 = 1/4


V = v/2

Second equation:


(1/2)M*(V + 1.4)^2 = (1/2)m*v^2


= (1/2)*(M/3)*(3V)^2


cancel out the M's and (1/2)'s


(V + 1.4)^2 = 3V^2


V^2 + 2.8V + 1.96 = 3V^2


V^2 -1.4V -0.98 = 0

V^2 = 0.98/0.4 = 2.45

V = 1.57

3 0
2 years ago
Read 2 more answers
A girl pulls her younger brother on a sled along a flat sidewalk (the total mass of the sled is 30kg). A frictional force of 50N
harkovskaia [24]

Answer:

6.13 s

219 N

Explanation:

Newton's law in the x direction:

∑F = ma

150 cos 30° N − 50 N = (30 kg) a

a = 2.66 m/s²

Δx = v₀ t + ½ at²

(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²

t = 6.13 s

Newton's law in the y direction:

∑F = ma

Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0

Fn = 219 N

4 0
3 years ago
How are mass, distance and gpe related
Gelneren [198K]

Answer:

It is direct proportionality. The greater the mass, the greater is the gravitational potential energy. The equation for GPE is : GPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the ground. As you can see GPE is directly proportional to mass, and height. KT.

Explanation:

Gravitational potential energy is a function of both the mass of your system and the mass of the thing generating the gravity field around your system.

The relationship is linear, which means that if you multiply or divide one of the masses by some number but leave everything else the same, you multiply or divide the potential energy by the same number. A 3kg mass has three times the gravitation potential energy of a 1kg mass, if placed in the same location.

6 0
2 years ago
A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
3 years ago
a man stands in a lift going downward with uniform velocity. he experiences a loss of weight at the start but not when lift is i
AnnyKZ [126]

Answer:

It is explained in the explanation section

Explanation:

When the lift starts going downwards, it will start accelerating downwards. After a while, it will start moving with a constant velocity.

Constant velocity means that acceleration is zero and so the man will not feel any weight loss.

Now, Once the lift achieves constant velocity the acceleration is zero hence he will not experience any weight loss.

However, when the lift is in uniform motion, the lift and the man will fall down with an acceleration(a) that is less than that due to gravity(g) . Thus, the man will feel an apparent weight F which is not equal to zero.

6 0
3 years ago
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