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3 years ago

15

If a car is moving on a straight line with a velocity of 40 m/s and it changes its velocity to 60 m/s in 4 seconds, calculate it

s acceleration.

1 answer:

Bogdan [553]3 years ago

8 0

Answer:

5m/s²

Explanation:

Initial Velocity= 40m/s

Final Velocity=60m/s

Time = 4s

a=(v-u)/t

a=60-40/4

a=20/4

a=5m/s²

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Take another look at lines 2 and 3. Suppose you use distance and time between any pair of neighboring dots to calculate speed:

Nataly [62]

Answer:

Add the two speeds together.

Then, divide the sum by two. This will give you the average speed for the entire trip. So, if Ben traveled 40 mph for 2 hours, then 60 mph for another 2 hours, his average speed is 50 mph.

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3 years ago

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Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

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3 years ago

The force exerted by the earth on a body is called ______of a body.

AVprozaik [17]

Answer:

The force exerted by the earth on a body is called gravitational of a body.

i think this is the answer

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3 years ago

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A 60 kg runner accelerates during a race at 3m/s2. what force is exerted on the earth with each step

marusya05 [52]

Answer:

180 Newton(N)

Explanation:

force =mass *acceleration

=60 * 3

=180 kgm/s^2

=180 N

6 0

3 years ago

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

MaRussiya [10]

Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = $\sqrt{ \frac{T}{\mu} }$

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = $\sqrt{ \frac{150}{0.0232} }$ = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s

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3 years ago