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3 years ago

11

Describe the motion above and answer

1 answer:

amm18123 years ago

7 0

Answer:

mmmmmmmm

Explanation:

mmmmmmmmmmmmmmmmmmmm

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Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= $\sqrt{(8)^2+(-2)^2}$

|v|= $\sqrt{68}$ = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0

2 years ago

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

= 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

= $0.5 x 25 x 5.6^{2}$ = 392 J

(C) average frictional force = $\frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}$

change in KE (ΔKE) = initial KE - final KE
ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$
when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

\frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

= $\frac{(-392 + 2940)}{12}$ = 212.33 N

5 0

2 years ago

Name three ways in which an object can accelerate

shutvik [7]

-Slow down
-Speed up
-Turning
Hope this helps

8 0

2 years ago

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

givi [52]

For t1:

t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec

For t2:

t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec

Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)

d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m

Where:

d = hor. distance

ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25

The answer is 3.19 : 2.25

8 0

3 years ago

Read 2 more answers

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its

True [87]

Answer:

The angular velocity is $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

The mass of wooden gate is $m_g = 4.5 kg$

The length of side is L = 2 m

The mass of the raven is $m_r = 1.2 kg$

The initial speed of the raven is $u_r = 5.0m/s$

The final speed of the raven is $v_r = 1.5 m/s$

From the law of conservation of angular momentum we express this question mathematically as

Total initial angular momentum of both the Raven and the Gate = The Final angular momentum of both the Raven and the Gate

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is zero

Note: This above is the generally formula for angular momentum of square objects

The final angular velocity of the Raven is $m_r * v_r * \frac{L}{2}$

The final angular velocity of the Gate is $\frac{1}{3} m_g L^2 w$

Substituting this formula

$m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

Substituting value

$\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

$6w = 4.2$

$w = \frac{6}{4.2}$

$w = 1.43\ rad/sec$

5 0

2 years ago