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2 years ago

Help me fill this out please (BRANLIEST!!!)

Physics

2 answers:

Dafna1 [17]2 years ago

7 0

Answer: For 18 the inch is longer im not sure by how much, for 19 the the kilogram is 2.2 times heavier than a pound, QUART IS A LITTLE LESS THAN A LITER. ... An easy way to figure from liters to gallons, for example, is that a quart is a little less than a liter and 4 liters is a little more than 1 gallon. To be exact, 1 liter is 0.264 gallon (a little more than a quart), and 4 liters is 1.06 gallons, 30°C or 30°F? Correct answer: the table shows that 30°C is about 86°F, which is warmer than 30°F. And sorry Im not sure how to do the rest

Explanation:

Nikitich [7]2 years ago

7 0

17.) 0.621 Mile
18.) Inch, 2.54 Centimeters
19.) Kilogram, 2.2 Pounds
20.) Liter, 1.057 Quarts
21.) Milliliter
22.) Celsius

Sorry couldn’t do 23 but I hope this helps! (^_^)

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A bowling ball of mass 5.8 kg moves in a

evablogger [386]

Answer:

Velocity of the ping pong ball must be = V2= 6,035.34m/s

Explanation:

M1= momentum of the bowling ball

m1 = mass of the bowling ball= 5.8kg

v1= velocity of the bowling ball= 1.59m/s

M2= momentum of the ping pong ball

m2= mass of the ping pong ball= 1.528 g/1000= 0.001528kg

v2= velocity of the ping pong ball

Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s

Momentum of the ping pong ball = M2= M1= m2v2

= 0.001528 *v2= 9.222

v2= 9.222/0.001528= 6,035.34 m/s

7 0

3 years ago

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-

Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

$v_{s_{x}}=0$

$v_{s_{y}}=2.0\ m/s$

The velocity of the boat in term x and y coordinate

For x component,

$v_{b_{x}}=v_{b}\cos\theta$

Put the value into the formula

$v_{b_{x}}=5.60\cos19$

$v_{b_{x}}=5.29\ m/s$

For y component,

$v_{b_{y}}=v_{b}\sin\theta$

Put the value into the formula

$v_{b_{y}}=5.60\sin19$

$v_{b_{y}}=1.82\ m/s$

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

$v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}$

Put the value into the formula

$v_{sb_{x}=0-5.29$

$v_{sb}_{x}=-5.29\ m/s$

For y component,

$v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}$

Put the value into the formula

$v_{sb_{x}=2.-1.82$

$v_{sb}_{x}=0.18\ m/s$

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

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olchik [2.2K]

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Explanation:

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