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3 years ago

A train travels 98 kilometers in 4 hours, and then 61 kilometers in 3 hours. What is its average speed?

1 answer:

6 0

Speed (velocity)=distance/time

V1=98km/4hr=24.5km/hr
V2=61km/3hr=20.3km/hr

Average speed (velocity)=total velocity/ number

Average speed (velocity)=44.8km/hr/2=22.4

So the average speed is 22.4km/hr

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When an electron moves 2.5 m in the direction of an electric field, the change in electrical potential energy of the electron is

anygoal [31]

In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,

E = V/d

However, the voltage in volts is energy per coulomb. Thus,
V = (8x10-17 J/electron)*(1electron/1.60218x10^-19 C)
V = 499.32 volts

Therefore,

E = 499.32 volts /2.5 m
E = 199.73 N/C

The electric field that caused the change in potential energy is equal to 199.73 Newtons per Coulomb.

7 0

3 years ago

Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav

DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

$v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s$

Time taken to reach 12% of light speed is 3673469.39 seconds

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m$

The distance it would have to travel is 6.61×10¹⁴ m

7 0

3 years ago

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp

IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0

3 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

3 years ago

Read 2 more answers

How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.

nata0808 [166]

Hello!

Answer:
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement.

We are going to use the following formula:

$W= F . Cos \alpha . D$

Where:

$F=214 N$
$\alpha =0$ º
$D= 37m$

Then, by substituting we have:

$W=214N . Cos (0).37m= 7918 N.m=7918 J$

8 0

3 years ago