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PilotLPTM [1.2K]
3 years ago
6

I need help with high school chemistry!!! Can someone help

Physics
1 answer:
Aleks04 [339]3 years ago
3 0
I can see what i can do if you need assistance
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If a sprinter’s mass is 60 kg, how much forward force must be exerted on the sprinter to make the sprinter accelerate at 0.8 m/s
Natasha2012 [34]
Forward force is 60×0.8
=48N
7 0
3 years ago
Read 2 more answers
How do you solve this ?<br> Why is the ans C not B ?
melomori [17]

Explanation:

Draw a free body diagram of the toolbox.  There are two forces:

Weight force mg pulling down,

and applied force F pulling up.

Sum of forces in the y direction:

∑F = ma

F − mg = ma

45 N − 15 N = (3 kg) a

a = 10 m/s²

The answer should be B.  It's possible the answer key has a mistake.

4 0
3 years ago
What do we mean when we say that 50 million tons of coal could replace the use of 0.6 MBPD (million barrels per day) of oil used
ololo11 [35]
<span>This statement simply implies that the 50 million tons of coal will produce the same energy output in one year as the 0.6 million barrels per day of oil. Assuming coal is more plentiful than oil, this is significant as 50 million tons is much smaller than 219 million barrels of oil.</span>
6 0
4 years ago
. The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum
Tju [1.3M]

Explanation:

120+60=180

120-60=60

5 0
3 years ago
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

 W=3420.84*0.25[2401-65536]

 W=17084965.5J

 W=17085KJ

 

'

'

4 0
3 years ago
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