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Aliun [14]
2 years ago
6

A. 24.89 B. 25.89 C. 17.74 D. 19.73

Physics
1 answer:
Veronika [31]2 years ago
5 0

Answer: D

Explanation:

Just did it got an 100

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CAN SOMEONE PLEASE TELL ME WHAT THIS WHEEL IS CALLED.WILL GIVE BRAINLIEST
kolbaska11 [484]

Answer:

its a pie chart

Explanation:

.........................

3 0
3 years ago
Read 2 more answers
Recalculate 100 km/h to m/s
qwelly [4]

= 27.777

Explanation:

A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.

4 0
3 years ago
A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin
Lelechka [254]

Answer:

Explanation:

Given

mass of boy m_b=70.9\ kg

mass of girl m_g=43.2\ kg

speed of girl after push v_g=4.64\ m/s

Suppose speed of boy after push is v_b

initially momentum of system is zero so final momentum is also zero because momentum is conserved

P_i=P=f

0=m_b\cdot v_b+m_g\cdot v_g

v_b=-\frac{m_g}{m_b}\times v_g

v_b=-\frac{43.2}{70.9}\times 4.64  

v_b=-2.82\ m/s

i.e. velocity of boy is 2.82 m/s towards west                

8 0
2 years ago
A 4kg brick is dropped from the top of a building whose hight is 30m.what is the velocity with which it reaches the ground​
miss Akunina [59]

Answer:

=24.25 ^−1

Explanation:

Let   and   be initial and final velocity of the body respectively,  

be acceleration due to gravity ( 9.8^−2 ),  ℎ be the height of the body.

=0 ^ −1

ℎ=30

we know that, ^2−^ 2=2ℎ

^2=2∗9.8∗30

^2=588

=24.25 ^−1

4 0
3 years ago
A 235.0 g metal block absorbs 2.44 × 103 J of heat to raise its temperature by 35 K. What is the specific heat of the metal? Sho
andrew-mc [135]
When an object absorbs an amount of energy equal to Q, its temperature raises by \Delta T following the formula
Q=m C_s \Delta T
where m is the mass of the object and C_s is the specific heat capacity of the material.

In our problem, we have Q=2.44 \cdot 10^3 J, m=235.0 g and \Delta T=35 K, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
C_s =  \frac{Q}{m \Delta T}= \frac{2.44 \cdot 10^3 J}{(235.0 g)(35 K)}=0.297 J g^{-1} K^{-1}
3 0
3 years ago
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