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1 year ago

I need help with problem C, finding the area of the accel. v time graph

Physics

1 answer:

erastova [34]1 year ago

7 0

c) To determine the velocity of the motion at the given time, consider that the velocity is also the area under the line of the graph a vs t.

Between t=0s and t = 2 s, the area under the line is:

A1 = 2 X 2 = 4

Then, until t = 2s the velocity is 4 m/s

The area between t = 2s and t = 4 s is:

A2 = 2 x 0 = 0

The area between t = 4s and t = 5s, the area is:

A3 = 2 x 1 = 2

Now, consider that the velocity is the sum of the previous areas, but taking areas on the negative y-axis as negative ones. Then, you have:

A = A1 + A2 - A3 = 4 + 0 - 2 = 2

Hence, the velocity is 2 m/s

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Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the

nignag [31]

Answer:

a = 5.5 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.

ΣF =m*a

where:

F = Force [N] (units of Newtons)

m = mass = 20 [kg]

a = acceleration [m/s²]

Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.

$230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ]$

5 0

3 years ago

You will get 50pts and the check mark.

Tamiku [17]

B) A ladybug crawling forward at constant rate of 2.5 m/s

6 0

3 years ago

Read 2 more answers

During a race, a dragster is 200m from the finish line when something goes wrong and it stops accelerating. It travels at a cons

GarryVolchara [31]

Answer:

135 m

Explanation:

Speed = 45 m/s, time = 3 second

The formula for the distance is given by

Distance = speed x time = 45 x 3 = 135 m

4 0

3 years ago

When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same

ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$ ...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

8 0

3 years ago

Read 2 more answers

Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electri

Studentka2010 [4]

Answer:

a) $E=2.88*10^{11}N/C$

b) $E=1.44*10^{11}N/C$

c) $F=4.61*10^{-8}N$

Explanation:

We use the definition of a electric field produced by a point charge:

$E=k*q/r^2$

a)Electric Field due to the alpha particle:

$E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C$

b)Electric Field due to electron:

$E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C$

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force but with opposite direction:

$F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N$

4 0

3 years ago