All categories

3 years ago

You are riding in a train that is traveling at

1 answer:

8 0

7 and a half hours because if you multiply 120 by 8 you get 960 and if you multiply 120 by 7 you get 840. So it’ll take 7 full hours and 30 minutes to reach 950 km. Hope this helps and I’m sorry if it’s wrong I did my best :)

You might be interested in

A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass

saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

$W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)$ (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

$d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m$

Furthermore, the angle α is:

$\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°$

Then, you solve the equation (1) for W and replace the values of all parameters:

$W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J$

The work done by the student is 609.97J

6 0

4 years ago

What is meant by pressure?write its unit.

Mars2501 [29]

Answer:

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed

Unit: Pascal (Pa)

1Pa = 1N/m^2

5 0

3 years ago

An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s^{2

erica [24]

(a) 18717 N

Newton's second law in this situation can be written as:

$\sum F = T-W = ma$ (1)

where

T is the tension in the cable, pointing upward

W is the weight of the elevator+passengers, pointing downward

m is the mass of the elevator+passengers (1700 kg)

a is the acceleration of the system (1.20 m/s^2, upward)

The weight is equal to the product between the mass, m, and the gravitational acceleration, g:

$W=mg=(1700 kg)(9.81 m/s^2)=16,677 N$

So now we can solve eq.(1) to find T, the tension in the cable:

$T=W+ma=16,677 N +(1700 kg)(1.20 m/s^2)=18,717 N$

(b) 16677 N

In this situation, the elevator is moving with constant velocity: this means that its acceleration is zero,

a = 0

So Newton's second law becomes

$\sum F = T-W = 0$

and so we find

$T=W=16,677 N$

(c) 15657 N

During the deceleration phase, Newton's second law can be written as:

$\sum F = T-W = ma$ (1)

Where the acceleration here points downward (because the elevator is decelerating), as the weight W, so we can write it as a negative number:

a = -0.600 m/s^2

we can solve the equation to find T, the tension in the cable:

$T=W+ma=16,677 N +(1700 kg)(-0.600 m/s^2)=15,657 N$

(d) 19.35 m, 0 m/s

Distance covered during the first part of the motion; we know that

u = 0 is the initial velocity

a = 1.20 m/s^2 is the acceleration

t = 1.50 s is the time

So the distance covered is given by

$d_1=ut + \frac{1}{2}at^2 = (0)(1.50 s)+\frac{1}{2}(1.20 m/s^2)(1.50 s)^2=1.35 m$

and the final velocity after this phase is

$v_1=u+at=0+(1.20 m/s^2)(1.50 s)=1.8 m/s$

During the 2nd part of the motion, the elevator moves at constant speed of 1.8 m/s for t=8.50 s, so the distance covered here is

$d_2 = v_1 t =(1.8 m/s)(8.50 s)=15.3 m$

Finally, in the third part the elevator decelerates at a = -0.600 m/s^2 for t = 3.00 s. So, the distance covered here is

$d_3 = v_1 t + \frac{1}{2}at^2=(1.8 m/s)(3.00 s) + \frac{1}{2}(-0.600 m/s^2)(3.00 s)^2=2.7 m$

and the final velocity is

$v_3 = v_1 +at = 1.8 m/s +(-0.600 m/s^2)(3.00 s)=0$

and the total distance covered is

$d=d_1 +d_2+d_3=1.35 m+15.30 m+2.70 m=19.35 m$

3 0

3 years ago

An object has the mass of 42g on earth, what is its mass on the moon

telo118 [61]

Answer: 42g

Mass will still stay the same because mass is a weight without pull of gravitational force. So whichever plant it is in, even in jupiter, the mass is going to be 42g.

7 0

3 years ago

14. Looking ahead: Give an example of how the Law of Conservation of Energy applies to eating your favorite meal. Give at least

Alexxandr [17]

Answer:

The law of conservation of energy applies to eating your favorite food in that the energy contained in the food is being converted to energy usable for the body system after the final digestion(Glucose or simple sugar, amino acid, etc)

Examples of transformation of energy include

(1) Transformation from Chemical energy in battery to light energy in bulb

(2) Transformation of Electrical energy to thermal/Heat energy in electric stove.

Explanation:

The law of conservation of energy states that energy can neither be created or distroy. It can only by converted from one form to another.

8 0

3 years ago