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3 years ago

PLS HELP ASAP :(( You observe a car parked on the side of the road. If the car started to move, Which conclusion could you make?

Physics

2 answers:

Nostrana [21]3 years ago

7 0

Answer:

The first option

Explanation:

Nadya [2.5K]3 years ago

5 0

Answer:

A (the first one)

Explanation:

An object is stationary when both forces acting on it are balanced. As soon as the car started moving, the driver had placed extra force on the accelerator and therfore the forces became unbalanced and the car moved

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A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0

3 years ago

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings

ozzi

Answer:

$0.3165\ \text{rad/s}$

Explanation:

m = Mass of person = 65 kg

d = Diameter of round table = 6.5 m

r = Radius = $\dfrac{d}{2}=3.25\ \text{m}$

v = Velocity of person running = 3.8 m/s

$I_t$ = Moment of inertia of turntable = $1850\ \text{kg m}^2$

Moment of inertia of the system is

$I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2$

As the angular momentum of the system is conserved we have

$L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}$

The angular velocity of the turntable is $0.3165\ \text{rad/s}$ .

3 0

3 years ago

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

amm1812

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

7 0

3 years ago

Read 2 more answers

- A cannon of 2000 kg fires a shell of 10 kg at

fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

$p=0$

After the shot, the momentum is:

$p=MV+mv$

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V, we find the velocity of the cannon:

$V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s$

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

$a=-0.25 m/s^2$ is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

$s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m$

7 0

3 years ago

The kinetic friction coefficient between a cabinet and the floor is 0.3. Mass of the cabinet is 300kg. A man pushes the cabinet

podryga [215]

Answer:

Explanation:

Step one:

given data

mass m= 300kg

applied force F= 1000N

coefficient of friction μ= 0.3

Step two:

The net force Fn= applied force-friction force

Fn=F-F1

F1= limiting force

F1=μ*m*g

F1=0.3*300*9.81

F1=882.9N

the Net force= 1000-882.9

Fn=117.1N

Step three:

we know that

F=ma

Fnet=ma

a= Fnet/m

a=117.1/300

a=0.39m/s^2

7 0

3 years ago