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3 years ago

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A rowboat heads directly across a river at a speed of 3 m/s. Select the correct equations that show that if the river flows at 4

m/s, the speed of the boat relative to the riverbank is 5 m/s.

1 answer:

olasank [31]3 years ago

6 0

Answer:

Explanation:

Check attachment for solution

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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

2 years ago

A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the

vitfil [10]

Beef and cheddar is your answer

3 0

2 years ago

What is the density of a block of wood with a mass of 120 g and a volume of 200 cm

creativ13 [48]

D=m/V therefore the answer is 120/200 or 0.6

7 0

2 years ago

If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

Leni [432]

Answer:

More force

Explanation:

Object A has more mass than object B

For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Net force = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

3 0

2 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago