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Doss [256]
3 years ago
12

A body accelerates by 25m/s 2 when it applied by 40n forces.what would be acceleration if it is applied by 80n forces

Physics
2 answers:
frutty [35]3 years ago
7 0

Answer:

50m/s2

Explanation:

since the force is doubled the acceleration will be doubled

il63 [147K]3 years ago
6 0

Answer:

50m/s²

Explanation:

cross muliplyy write down the values acceleration and force

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Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then
Andreas93 [3]

Answer:

The displacement is  \Delta H =    -  1 \ m

The distance  is  D =  4 \  m

Explanation:

From  the question we are told that

    The height from which the ball is dropped is  h  =  1 \ m

    The height attained at  the first bounce is  h_1  = 0.8  \  m

    The height attained at  the second bounce is   h_2 = 0.5 \  m

    The height attained at  the third bounce is h_3 = 0.2 \  m

Note  : When calculating displacement we consider the direction of motion

Generally given that upward is positive  the total displacement of the ball is mathematically represented as

            \Delta H =  (0  -  h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)

Here the 0 show that there was no bounce back to the point where Billy released the ball  

           \Delta H =  (0  -  1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)

=>          \Delta H =    -  1 \ m

Generally the distance covered by the ball is mathematically represented as  

                D =  h +  2h_2 + 2h_3 + 2h_3

The 2 shows that the ball traveled the height two times

              D =  1 +  2* 0.8  + 2* 0.5 + 2* 0.2

=>           D =  4 \  m

     

5 0
3 years ago
How do hormones affect only certain cells in the body but not others?
Snezhnost [94]
 I<span>f you have studied enzymes its a similar concept. Cells have proteins on the surface of their cell which hormones bind to (called receptors) The receptor must be a complimentary shape to the hormone for it to bind. Only target cells have the receptor with the complimentary shape so only these cells will be affected.</span>
4 0
3 years ago
Starting at 48th Street, Dylan rides his bike due east on Meridian Road with the wind at his back. He rides for 20 min at 15 mph
Nezavi [6.7K]

Answer:

55 min

Explanation:

The missing question is: how long does the trip take?

First of all, we need to find the initial distance covered by Dylan. In the first part, he rides for

t_1 = 20 min = \frac{1}{3}h

at a speed of

v = 15 mph

therefore, the distance he covered is

d = v t_1 = (15)(\frac{1}{3})=5 mi

Then Dylan stopped for a time of

t_2 = 5 min = \frac{5}{60}=\frac{1}{12}h

Finally, on the way back, Dylan covered again this distance but travelling at a new speed of

v = 10 mph

So, the time he took is

t_3 = \frac{d}{v}=\frac{5}{10}=\frac{1}{2}h = 30 min

So, the total time of the trip was

t=t_1 + t_2 + t_3 = 20 min + 5 min + 30 min = 55 min

6 0
3 years ago
A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2
saw5 [17]

Answer:

Half as large.

Explanation:

Using Newton's law of universal gravitation, if the mass of the planet is <em>M</em> and of the Moons 1 and 2 is <em>m</em>, them the force exerted by the planet on them will be:

F_1=\frac{GMm}{r}

F_2=\frac{GMm}{2r}

Which clearly shows that the force that the planet exerts on the Moon 2 is half  the force it exerts on the Moon 1.

7 0
3 years ago
A ____ is the time required for one half of the nuclei in a radio- ____ isotope to decay.
sukhopar [10]

Answer:

A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.

Explanation:

A radio-active isotope is an isotope which undergoes radioactive decay.

Radioactive decay is a spontaneous process in which the nucleus of an atom changes its state (turning into a different nucleus, or de-exciting), emitting radiation, which can be of three different types: alpha, beta or gamma.

The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.

The law of radio-active decay can be expressed as follows:

N(t) = N_0 (\frac{1}{2})^{t/t_{1/2}}

where

N(t) is the number of undecayed nuclei left at time t

N0 is the initial number of nuclei

t is the time

t_{1/2} is the half-life

We see that when t=t_{1/2} (that means, when 1 half-life has passed), the number of undecayed nuclei left is

N(t) = N_0 (\frac{1}{2})^{t_{1/2}/t_{1/2}}=N_0 (\frac{1}{2})^1=\frac{N_0}{2}

So, half of the initial nuclei.

5 0
3 years ago
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