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Novay_Z [31]
3 years ago
11

An ice boat is coasting along a frozen lake. Friction between the ice and the boat is negligible, and so is air resistance. Noth

ing is propelling the boat. From a bridge someone jumps straight down and lands in the boat, which continues to coast straight ahead. (a) Does the horizontal momentum of the boat change? (b) Does the speed of the boat increase, decrease, or remain the same? Explain.
Physics
1 answer:
zhannawk [14.2K]3 years ago
6 0

Answer:

(A) No

(B) Speed decreases

Explanation:

(A) since there is nothing propelling the boat and the friction between the ice and the boat and also air resistance is negligible the net force of the system in the horizontal direction is zero and hence there is no change in the horizontal momentum of the boat.

(B) Since the person had not velocity in the horizontal direction before landing on the boat but now has one after landing on the boat, the speed of the boat will decrease because the momentum has to be conserved (remember there is no change in it).

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Katrina’s recipe for orange muffins calls for 400 grams of flour. How many kilograms of flour are in the recipe?
aliina [53]

Answer:

0.4 Kilograms of flour

Explanation:

If you convert 400 grams into kilograms (400 ÷ 1,000), you would get 0.4 kilograms.

3 0
3 years ago
Pls give me an answer
wolverine [178]

Answer:

IV what is it's potential energy at the maximum height

4 0
3 years ago
Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k
Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is \bf{6.59~m/s}.

Part(B): The direction is \bf{42.7^{0}} south to east.

Explanation:

Given:

Mass of Jack, m_{1} = 59.0~Kg

Mass of Jill, m_{2} = 47..0~Kg

Initial velocity of Jack, v_{1i} = 8.00~m/s

Initial velocity of Jill, v_{2i} = 0

Final velocity of Jack, v_{1f}  5.00~m/s

The final angle made by Jack after collision, \alpha = 34.0^{0}

Consider that the final velocity of Jill be v_{2f} and it makes an angle of \beta with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}

where, v_{2f}^{x} is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s

Conservation of momentum of the system along north direction is given by

~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s

where, v_{2f}^{y} is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s

Part(B):

The direction is given by

\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}

4 0
3 years ago
Find the volume of a box with a length of 5 cm, a width pf 5cm, and a height of 10cm.
NNADVOKAT [17]

Answer:

250 cm³

Explanation:

given,

Length of the Box, L = 5 cm

Width of Box, W = 5 cm

height of the box, H = 10 cm

Volume of the Box = L W H

V = 5 x 5 x 10                    

V = 250 cm³                            

Volume of the box is equal to 250 cm³

7 0
3 years ago
A particle accelerates from rest at 5.7 m/s 2 . what is its speed 7.2 s after the particle starts moving? answer in units of m/s
True [87]
It starts from rest meaning the initial velocity is 0

is accelerates at 5.7m/s^2 meaning a = 5.7m/s^2

t = 7.2s

vf = vi + at 

vf = final velocity
vi = initial velocity 
a = acceleration 
t = time 

we are looking for the final velocity 

vf = vi + at 

vf = 0 + (5.7 * 7.2)

vf = 41.04 m/s  -------This is your answer


6 0
3 years ago
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