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2 years ago

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a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis

placement of the bus. (Answer is 4.7x10^2m)

1 answer:

KatRina [158]2 years ago

7 0

The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.

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A ship sends out a 1200 Hz sound wave, which has a wavelength of 120 cm in the water. What would happen if that ship sent out a

OlgaM077 [116]

Answer:

the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

Explanation:

From wave,

v = λf................ Equation 1

Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: f = 1200 Hz, λ = 120 cm = 1.2 m

Substitute into equation 1

v = 1200(1.2)

v = 1440 m/s.

When the ship sent out a 600 Hz sound wave,

make λ the subject of formula in equation 1

λ = v/f............. Equation 2

Given: f = 600 Hz, v = 1440 m/s

Substitute into equation 2

λ = 1440/600

λ = 2.4 m or 240 cm.

When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

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2 years ago

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

likoan [24]

Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

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2 years ago

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

$V_1=V_2=50912\ V$

The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

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2 years ago

Two rocks are tied to massless strings and whirled in nearly horizontal circles so that the time to travel around the circle onc

Fynjy0 [20]

Answer: $m_1=m_2$

Explanation:

Given

Time period for both string is same

$\frac{2\pi r}{v_1}=\frac{2\pi 2r}{v_2}$

$2v_1=v_2$

and tension in string 2 is twice the first string

$2T_1=T_2$

Tension will provide centripetal acceleration

$2\frac{m_1v_1^2}{r}=\frac{m_2v_2^2}{2r}$

$2\frac{m_1v_1^2}{r}=\frac{m_2\times 4v_1}{2r}$

thus $m_1=m_2$

3 0

2 years ago

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

iragen [17]

Answer:

The workdone is $W_d =-4400J$

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

The force is on the force gauge $F = 2750 N$

The distance that Magnus pulled the bus $d = 1.60m$

Generally the workdone by the tension force on Magnus is

$Workdone = Force * displacement \ in \ the \ direction \ of \ force$

$W_d = F * (-d)$

This negative sign show that is tension force is in the opposite direction to Magnus movement (i.e the movement of the bus )

Substituting value we have

$Workdone = - 2750 * 1.60$

$=-4400 J$

7 0

2 years ago