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2 years ago

12

A car is traveling with speed v0 when it begins to speed up at a rate of Δv every second. After t1 seconds, the car travels with

zero acceleration for t2 seconds. Which of the following is a correct expression for the displacement of the car during this motion?

1 answer:

Rainbow [258]2 years ago

8 0

Answer:

d = Δv(t2-t1)

Explanation:

Speed is defined as the change of displacement with respect to time. It is expressed as shown;

Speed = change in displacement/change in time

Δv = d/Δt

d = Δv*Δt

d = ΔvΔt

Δt = t2-t1

d = Δv(t2-t1)

Δv is the change in rate of speed

Δt = change in time

The correct expression for the displacement of the car during this motion is d = Δv(t2-t1)

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A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1

const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0

2 years ago

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

6 0

2 years ago

Read 2 more answers

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

$\lambda = 0.186 m$

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

$f' = f_0 (\frac{v + v_o}{v - v_s})$

here we know that

$f_0 = 1750 Hz$

$v_o = 15 m/s$

$v_s = 15 m/s$

now we will have

$f' = (1750)(\frac{340 + 15}{340 - 15})$

$f' = 1911.5 Hz$

Part b)

Apparent wavelength is given by the formula

$\lambda = \frac{v_{relative}}{f_{app}}$

here we will have

$\lambda = \frac{340 + 15}{1911.5}$

$\lambda = 0.186 m$

3 0

2 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

2 years ago

Ball 1 travels with a momentum of 48.0 kg-m/s east and strikes Ball 2, which is initially at rest.. Ball 1 separates at an angle

vladimir1956 [14]

Answer:

Momentum of 2nd ball is

$P = 31.6 kg m/s$

direction is given as

$\theta = -37.66 degree$

Explanation:

As we know that there is no external force on the system of balls so momentum before and after collision will be conserved

So we have

$P_i = 48 \hat i + 0$

now after collision momentum of two balls is must be same as initial

so we have

$P_i = P_f$

$48\hat i = (30 cos40 \hat i + 30 sin40\hat j) + (P_{2x}\hat i + P_{2y}\hat j)$

so we have

$48 = 23 + P_{2x}$

$P_{2x} = 25 kg m/s$

for other component we have

$0 = 19.3 + P_{2y}$

$P_{2y} = -19.3 kg m/s$

Momentum of 2nd ball is given as

$P = \sqrt{P_2x}^2 + P_{2y}^2}$

$P = 31.6 kg m/s$

direction is given as

$tan\theta = \frac{P_{2y}}{P_{2x}}$

$tan\theta = \frac{-19.3}{25}$

$\theta = -37.66 degree$

5 0

2 years ago