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Temka [501]
3 years ago
8

A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera

te foward. The initial acceleration of the train is 0.75m/s. Calculate the resultant force that acts on train this time​.
Physics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

420000N

Explanation:

Given parameters:

Mass of the train  = 5.6 x 10⁵kg

Acceleration  = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

   Force  = mass x acceleration

Resultant force that acts on the train is given below;

 Force  = 5.6 x 10⁵kg  x 0.75m/s²  = 420000N

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anzhelika [568]

Answer:

1⁺ ion

Explanation:

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Most metals are electropositive and would prefer to lose electrons than to gain it.

Like all metals, the group 1 elements called the alkali metals would prefer to lose and electron.

On losing an electron the number of protons is then greater than the number of electrons. This leaves a net positive charge.

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Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
antoniya [11.8K]

Answer:

r=5.278\times 10^{-4}\ m

Explanation:

Given that:

  • magnetic field intensity, B=0.07\ T
  • kinetic energy of electron, KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J
  • we have mass of electron, m=9.1\times 10^{-31}\ kg

<em>Now, form the mathematical expression of Kinetic Energy:</em>

KE= \frac{1}{2} m.v^2

1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2

v^2=4.2198\times 10^{11}

v=6.496\times 10^6\ m.s^{-1}

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

r=\frac{m.v}{q.B}

r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}

r=5.278\times 10^{-4}\ m

6 0
3 years ago
There are 2.2 pounds in a kilogram. if a boys mass is 40kg, what is his height in pounds?
sasho [114]
The simplest way to do this is to set up equivalent fractions, like this- 

\frac{1}{2.2} = \frac{40}{x}

Solve for x by using cross multiplication.

40*2.2= 88
1*x=88
x=88

Therefore, the boy weighs 88lbs. 
3 0
3 years ago
Read 2 more answers
The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

t = \frac{d}{v_{s} }

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be v_{r} = 2.5 m/s

S = v_{r} t

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s

The vertical displacement is given by, y = v_{y} t

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

7 0
3 years ago
Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

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e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
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