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vekshin1
3 years ago
14

1. Using pollen grains placed in water,

Physics
1 answer:
Bezzdna [24]3 years ago
6 0

hope it helps!!!!!!!!!!!!!!!

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An object travels at a speed of 7500 cm/sec . how far will it travel in kilometers in one day
DiKsa [7]

Answer:

6480 km

Explanation:

The speed of the object is

v = 7500 cm/sec

We need to convert centimetres into kilometers and seconds into days. We have:

1 cm = 1\cdot 10^5 km

1 s = \frac{1}{60\cdot 60 \cdot 24}d

Using these conversion factors, we find:

v=7500 \frac{cm}{s} \cdot 1\cdot 10^{-5} \frac{km}{cm}\cdot (24)(60)(60) \frac{s}{d}=6480 km

3 0
4 years ago
An internal explosion breaks an object, initially at rest, into two pieces: A and B. Piece A has 1.9 times the mass of piece B.
Helen [10]

Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

<h3>What is the relation between the masses of A and B?</h3>
  • Let mass of piece A = Ma

Mass of piece B = Mb

  • Velocities of pieces A and B are Va and Vb respectively.
  • As per conservation of momentum,

Ma×Va = Mb×Vb

  • Here, Ma=1.9Mb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

<h3>What are the kinetic energy of piece A and B?</h3>
  • Expression of kinetic energy of piece A = 1/2 × Ma × Va²
  • Kinetic energy of piece B = 1/2 × Mb × Vb²
  • Total kinetic energy= 7900J

=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

  • Kinetic energy of piece B = 7900 - 2724 = 5176 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ1

5 0
2 years ago
When applied to ecology,what do the new science of chaos imply
gladu [14]

equilibrium i think if not sorry

3 0
3 years ago
Some children are underfed and often have no place to sleep. According to
jeka94

Answer:

Physiological needs

Explanation:

3 0
3 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
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