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3 years ago

HELPP WILL MARK BRAINLIEST!!!

Physics

2 answers:

seraphim [82]3 years ago

4 0

Answer: the answer is c

guapka [62]3 years ago

3 0

I guess it’s B cause that maybe is the output

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The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.

andrew-mc [135]

Answer:

answer is 3.05v

Explanation:

hope it is helpful and briliant

6 0

2 years ago

How would you define physics class in your own words ?

Andrei [34K]

Answer:

honestly i dont like physics class but for you im gonna write somethin' good but for me tho its B O R I N G

Explanation:

Physics is the branch of science that deals with the structure of matter and how the fundamental constituents of the universe interact. It studies objects ranging from the very small using quantum mechanics to the entire universe using general relativity.

3 0

3 years ago

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th

Kryger [21]

Answer:

$H = 10.05 m$

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

$\Delta t = 4.1 - 0.5 = 3.6 s$

now we have

$\Delta t = \frac{2v}{g}$

$3.6 = \frac{2v}{9.8}$

$v = 17.64 m/s$

so this is the speed at the top of flag pole

now we have

$v_f - v_i = at$

$17.64 - v_i = (-9.8)(0.5)$

$v_i = 22.5 m/s$

now the height of flag pole is given as

$H = \frac{v_f + v_i}{2}t$

$H = \frac{22.5 + 17.64}{2} (0.5)$

$H = 10.05 m$

5 0

3 years ago

Pls I need help for the problem solving part.

Bogdan [553]

i don't know the anss , sorry.

3 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

3 years ago