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3 years ago

A train moves at constant velocity of 50km/h. how far will it move in 0.5h?

1 answer:

7 0

The formula for getting the distance will be distance = speed x time
                                                                               D = S x T
   speed or velocity = 50km/h
   time = 0.5 h
the equation will be done directly because it's already in it's SI units
            distance = 50km/h x 0.5h
            hour cancels hour and the equation remains = 50km x 0.5
                               Ans = 25 km
the train will move 25 km far in 0.5h

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Number of atoms of each element potassium chlorate?

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5 atoms is the answer

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2 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

$V_{x} = 11. 78 m/s$

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

= 37.303 + 85.994

= 123. 297 m

≈ 123. 30 m

4 0

2 years ago

Read 2 more answers

An AC generator consists of eight turns of wire, each of area 0.0775 m2 , and total resistance of 8.53 Ω. The loop rotates in th

Bad White [126]

Answer:

44.08 Volt

Explanation:

N = 8, A = 0.0775 m^2, R = 8.53 ohm, B = 0.222 T, f = 51 Hz

e0 = N B A w

e0 = 8 x 0.222 x 0.0775 x 2 x 3.14 x 51

e0 = 44.08 Volt

3 0

3 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

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3 years ago

A car is moving in uniform circular motion. If the car's speed were to double to keep the car moving with the same radius, the a

Anestetic [448]

Answer:

Increase by a factor of 4.

Explanation:

The acceleration of a car moving with speed v in a circle of radius R is given by:

$a=\frac{v^2}{R}.$

Now if we double the speed $v$ in the equation above, it becomes $2v$ . Thus:

$a=\frac{v^2}{R}\:\: {\rightarrow}\:\: a_n=\frac{(2v)^2}{R}=4\frac{v^2}{R}=4a.$

Therefore the acceleration is increased by a factor of 4.

7 0

3 years ago