What is the acceleration of a 59 kg object pushed with a force of 500 Newton's ?

A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumpe

Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

$U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where we have

$k=2\cdot 10^6 N/m$ is the spring constant of the bumper

x is the maximum compression of the bumper

$m=30\cdot 10^4 kg$ is the mass of the car

$v=2.0 m/s$ is the speed of the car

Solving for x, we find the maximum compression of the spring:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m$

8 0

3 years ago

Read 2 more answers

PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)

Luden [163]

Answer:

1) a radio are uses by astronomy

2) 6 bilion waves

3) expert vertified

Explanation:

1) in contrast to an "ordinary" telescope, which receives visible light, a radio telescope "sees" radio waves emitted by radio sources, typically by means of a large parabolic ("dish") antenna, or arrays of them.and Radio telescopes are also the primary means to track space probes, and are used in the SETI project. so must been

radio are almostly

ceiver by astronomy

2) Radio waves have the longest wavelengths in the electromagnetic spectrum.

3)Expert Verified

Radio telescopes are telescopes that are specially designed for observation of long light wavelengths

CARRY ON ✨

3 0

3 years ago

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$

v = $\frac{94.30}{120} \ 10^3$

v = 785.9 m / s

in the opposite direction, that is, the angle must be

41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

To find the initial Barry trajectory is 94.30 10³ m with n angles of 138.8º

The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here: brainly.com/question/15074838

4 0

2 years ago

A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove

jonny [76]

We must remember that the total net force equation at constant velocity is:

F – Ff = 0

of

F - µN = 0

Using Newton's 2nd Law of Motion:

F = m a

Where,

F = net force acting on the body
m = mass of the body
a = acceleration of the body

Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to

F = net Force = 0

Therefore,

F - µN = 0

where

µ = coefficient of friction = 0.20
N = normal force acting on the cart = 12 N

Therefore,

F - 0.20(12) = 0

F = 2.4 N

4 0

3 years ago

What mass of steam at 100°C must be mixed with 119 g of ice at its melting point, in a thermally insulated container, to produce

N76 [4]

Answer: $M=27.92\ gm$