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skelet666 [1.2K]
3 years ago
15

a roller coaster moves on a certain section of it's track with an average speed of 13 m/s. how much distance does it cover in 5.

8 s?​
Physics
2 answers:
viva [34]3 years ago
7 0
  • Speed=13m/s
  • Time=5.8s

\\ \sf\longmapsto Speed=\dfrac{Distance}{Time}

\\ \sf\longmapsto Distance=Speed\times Time

\\ \sf\longmapsto Distance=13(5.8)

\\ \sf\longmapsto Distance=75.4m

Yuliya22 [10]3 years ago
4 0

Answer:

<h2><em>75.4 m</em></h2>

Explanation:

<h3>Given,</h3>

Speed = 13 m/s

Time = 5.8 seconds

<h3>We know,</h3>

speed =  \frac{d}{t}

So therefore,

distance = s \times t

=  > distance = 13 \times 5.8

=  > distance = 75.4

As in the question it is mentioned m/s

The unit of length will be in meter

i.e

<h2>75.4 m</h2>

HOPE IT HELPS YOU!!

#IndianMurga

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Keith_Richards [23]

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

F = m\cdot(g-a)

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}

So

F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN

3 0
2 years ago
Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h
Stels [109]

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion \alpha= 10.5\times10^{-6}\ K^{-1}

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

\Delta L=L\alpha\Delta T

Where, \Delta T= temperature difference

\alpha=Coefficient of thermal expansion

L= length

Put the value into the formula

\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})

\Delta L=0.42\ m

Hence, The difference in the length of the bridge is 0.42 m.

5 0
3 years ago
Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of
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Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

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6 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

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F = m·a

where:

F = force

m = mass of the object

a = acceleration

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F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
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3 years ago
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