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Akimi4 [234]
4 years ago
11

A 777-\text{ kg} kgstart text, space, k, g, end text object is accelerating to the right at 2 \text{ km}/\text{s}^22 km/s 2 2, s

tart text, space, k, m, end text, slash, start text, s, end text, squared. What is the magnitude of the rightward net force acting on it
Physics
1 answer:
Pachacha [2.7K]4 years ago
3 0

Answer: 14000 N

Explanation: i just did it on khan academy

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Which trial’s cart has the greatest momentum at the bottom of the ramp?
Licemer1 [7]
The momentum of each cart is given by:
p=mv
where
m is the mass of the cart
v is its velocity (at the bottom of the ramp)

To answer the problem, let's calculate the momentum of each of the 4 carts:
1) p=(200 kg)(6.5 m/s)=1300 kg m/s
2) p=(220 kg)(5.0 m/s)=1100 kg m/s
3) p=(240 kg)(6.4 m/s)= 1536 kg m/s
4) p=(260 kg)(4.8 m/s)=1248 kg m/s

Therefore, the cart with greatest momentum is cart 3, so the right answer is
<span>- trial 3, because this trial has a large mass and a large velocity</span>
8 0
3 years ago
A 22kg box slides down a 30.0 degree incline. If there is no friction, what is the acceleration of the box?
Rashid [163]

Answer:

Acceleration of the box is 4.9 m/s².

Explanation:

The free body diagram is shown below.

The weight of the body can be resolved into two mutually perpendicular components as shown. The force responsible for sliding motion of the box along the incline is due to mg\sin \theta acting along the incline.

Therefore, as per Newton's second law of motion,

Net force = mass\times acceleration

Now, net force acting along the incline is only mg\sin \theta

Therefore,

mg\sin \theta=ma\\a=g\sin \theta

Now, g=9.8\textrm{ and } \theta = 30

Therefore, the acceleration of the box is given as:

a=9.8\times \sin(30)=9.8\times 0.5=4.9\textrm{ }m/s^{2}

Acceleration of the box is 4.9 m/s².

6 0
4 years ago
A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its equilibrium position, wha
Paha777 [63]

Answer:

3.At equilibrium, its instantaneous velocity is at maximum

Explanation:

The motion of a mass on the end of a spring is a simple harmonic motion. In a simple harmonic motion, the total mechanical energy of the system is constant, and it is sum of the elastic potential energy (U) and the kinetic energy of the mass (K):

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 = const.

where

k is the spring constant

x is the displacement of the spring from equilibrium

m is the mass

v is the speed

As we see from the formula, since the total energy E is constant, when the displacement (x) increases, the speed (v) increases, and viceversa. Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.

7 0
3 years ago
a 10 kg solid disk of radius 0.5 m is rotated about an axis through its center.the disk accelerates from rest to angular speed o
mezya [45]

Answer:

τ ≈ 0.90 N•m

F =

Explanation:

I = ½mR² = ½(10)0.5² = 1.25 kg•m²

α = ω²/2θ = 3.0² / 4π = 0.716... rad/s²

τ = Iα = 1.25(0.716) = 0.8952... ≈ 0.90 N•m

τ = FR

Now we have the unanswered question of reference frame.

80° from what?

If it's 80° from the radial

F = τ/Rsinθ = 0.90/0.5sin80 = 1.818...  ≈ 1.8 N

If it's If it's 80° from the tangential

F = τ/Rcosθ = 0.90/0.5cos80 = 10.311...  ≈ 10 N

There are an infinite number of other potential solutions

7 0
3 years ago
2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral
AnnyKZ [126]

Answer:

b. negative

Explanation:

neutrons have a negative charge and protons have a proton has a positive charge

3 0
3 years ago
Read 2 more answers
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