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Stells [14]
2 years ago
11

How much protein would be consumed with a 3-ounce serving of

Physics
1 answer:
dalvyx [7]2 years ago
6 0

Answer:

15kg

(i actually did this)

Explanation:

k12

hope this helped!

i hope you ace it!

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What would be an example of a chemical change that took a long time to occur
weqwewe [10]
<span>rusting takes a long time, and rusting is a chemical change</span>
3 0
3 years ago
Read 2 more answers
A racing car accelerates at the end of the race from a speed of 100
nevsk [136]

4/3 m/s ( approximately 1.3333... m/s)

5 0
2 years ago
A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +16.7 m/s in a time of 20.7 s.
frutty [35]

Answer:

F = 479.21 N

Explanation:

given,

initial velocity = 0 m/s  

final velocity = 16.7 m/s        

time taken = 20.7 s          

combined mass of the boat and trailer  = 594 kg

tension in the hitch = ?          

using equation of motion          

v = u + a t

16.7 = 0 + a × 20.7

a = 0.807 m/s²              

Force = mass × acceleration        

F = 594 × 0.807                

F = 479.21 N

Hence, the tension in the hitch that connects the  trailer to the car is        F = 479.21 N

7 0
3 years ago
An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If
wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by  850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0
3 years ago
An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera
Ede4ka [16]
The amount of heat given by the water to the block of ice can be calculated by using
Q=m_w C_{sw} \Delta T_w
where 
m_w = 30 g is the mass of the water
C_{sw}=4.18 J/(g ^{\circ}C) is the specific heat capacity of water
\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C is the variation of temperature of the water.

Using these numbers, we find
Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
Q = m_i L_f
where m_i is the mass of the ice while L_f =334 J/g is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
m_i =  \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g
3 0
2 years ago
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