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3 years ago

What vaciole-like strucyure break down and recycle cell parts

Physics

1 answer:

kap26 [50]3 years ago

6 0

<span>lysosomes should be the right answer to this question</span>

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Explain why electromagnetic wave is transverse wave and not a compressional wave. ...?

tiny-mole [99]

An electromagnetic , transverse, wave is a wave that goes up and down, such as light that does not need a medium to travel
While compressional, longitudinal wave such as sound require a medium to travel and can be compressed

hope this helps

5 0

3 years ago

How much work is done in holding a 15 N sack of potatoes while waiting in line at the

marta [7]

Answer: 2700 Joules

Explanation:

Given variables are:

Workdone (w) = ?

Weight of sack (w) = 15N

Time taken (t) = 3 minutes

The standard unit of time is seconds, so convert 3 minutes to seconds

(If 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Recall that work is done is when force is applied on an object over a distance

i.e Workdone = force x distance.

(However, the force acting on the sack of potatoes is weight)

Hence, workdone = weight x time

w = 15N x 180 seconds

w = 2700 J

thus, 2700 joules of work was done.

4 0

4 years ago

from the edge of the rooftop of a building, a boy thros a stone at an angle of 25 above the horizontal. The stone hits the groun

ella [17]

Answer:38.675 m/s

Explanation:

$\theta =25^{\circ}$

t=4.2 s

$R_x=105m$

Solving in x direction

$R_x=ut+\frac{1}{2}\left ( 0\right )t^2$

$105=ucos25\times 4.2$

ucos25=25

u=27.584 m/s

Initial velocity in vertical direction usin25

Let h be the height of Rooftop

$h=\left ( -usin25\right )4.2+\frac{1}{2} \left ( 9.8\right )\left ( 4.2\right )^2$

h=37.47 m

Therefore final vertical velocity is $v_{vf}^2-\left ( usin25\right )^2=2\times 9.81\times 37.47$

$v_{vf}$ =29.51 m/s

Final Resultant velocity

$V_{final}=\sqrt{29.51^2+24.99^2}$

$V_{final}=38.675 m/s$

6 0

4 years ago

The buoyant force acts in all directions. Please select the best answer from the choices provided T F

tester [92]

Where are the answer from the choices ?

5 0

3 years ago

Steam enters an adiabatic turbine at 5 MPa and 4500C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbi

REY [17]

Answer:350.92 KJ/kg

Explanation:

Given the process is reversible adiabatic i.e it is isentropic

$P_1=5 MPa$

$T_1=4500 MPa$

$P_2=1.4 MPa$

From steam table

$h_1=3317.03KJ/kg$

For isentropic process $s_1=s_2$

at $P_2=1.4 MPa$

$s_2=6.82 KJ/kg-k$

$h_2=2966.11 KJ/kg$

Therefore Work output of the turbine per unit mass of steam is = $h_1-h_2$

=3317.03-2966.11

=350.92 KJ/kg

8 0

3 years ago