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AlladinOne [14]
2 years ago
8

A molecule that moves across a cell membrane without using the cell’s energy tends to move

Physics
1 answer:
zaharov [31]2 years ago
4 0

Answer:answer is b

Explanation:

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Please help me with questions 1, 2 and 3. <br> i need a step by step explanation
kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) x=\frac{1}{2}at^2+v_0t+x_0

The equation for velocity v and a constant acceleration a is:

(ii) v=at+v_0

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) a = \frac{v -v_0}{t}

(iv) x = \frac{v-v_0}{2t}t^2+v_0t + x_0

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}

5 0
3 years ago
Does wind have atoms and molecules in it?
Likurg_2 [28]
<span>
Of course.  Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc.  That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>


4 0
3 years ago
Read 2 more answers
A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the
____ [38]
 <span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 

it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
--------------------------------------... 
let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
u^2 sin^2 p/2g = 1.5 
u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
===================== 
t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component 
check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 

so it will able to cross</span>
8 0
2 years ago
The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect
Diano4ka-milaya [45]
Rx= 3.5 km

Ry= 2.9 km
4 0
3 years ago
Read 2 more answers
a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition
MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

5 0
2 years ago
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