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3 years ago

6

As you go up a food web is energy lost or gained? A/ lost b/ gained

2 answers:

Hatshy [7]3 years ago

8 0

I'm preeeety sure the answer is A, lost! Hope I helped, hon :)

Oliga [24]3 years ago

8 0

The answer is A hope it helps

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At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball

Ymorist [56]

Answer:440.03 N

Explanation:

Given

horizontal component of acceleration $(a_x)=800 m/s^2$

vertical component of acceleration $(a_y)=880 m/s^2$

mass of ball =0.37 kg

Force in horizontal direction $=m\times a_x=0.37\times 800=296 N$

Force in vertical direction $=m\times a_y=0.37\times 880=325.6 N$

Therefore net force is

$F=F_x+F_y$

$|F|=\sqrt{296^2+325.6^2}$

|F|=440.03 N

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3 years ago

What is the total amount of force needed to keep a 6.0 kg object moving at speed

DanielleElmas [232]

Answer:

C. 0 N

Explanation:

In the absence of external forces, a body in motion will stay in motion.

F = ma

F = 6.0(0) = 0

8 0

3 years ago

Read 2 more answers

A reconnaissance plane flies 545 km away from

Nikolay [14]

Answer:

681.6/ms

Explanation:

A reconnaissance plane flies 545 km away from its base at 568 m/s. then flies back to its base at 852 m/s.

What is its average speed?

Answer in its of m/s

Avg speed of the round trip is

2*568*852/(568+852)= 681.6/ms

6 0

3 years ago

A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli

Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

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3 years ago

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

$Q_2=123-195=-72 J$

overall heat added $=Q_1+Q_2$

$Q_{net}=72-72 =0$

For overall Process Heat added is 0 J

8 0

3 years ago