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Aleksandr [31]
3 years ago
13

A 3.0-kilogram object is acted upon by an impulse having a magnitude of 15 newton•seconds. What is the magnitude of the object’s

change in momentum due to this impulse?
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0
The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.
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An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo
saveliy_v [14]

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

h = ut + \frac{1}{2}at^{2}

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

3 0
3 years ago
What is the volume of a cube with length = 3 centimeters, width = 3
suter [353]

Answer:

D. 27 cubic centimeters

Explanation:

l*w*h

3*3*3= 27

4 0
3 years ago
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
How many infrared photons of frequency 2.57 x 1013 Hz would need to be absorbed simultaneously by a tightly bound molecule to br
victus00 [196]

Answer:

94

Explanation:

f = 2.57 x 10^13 Hz

E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J

Energy of each photon = h f

Where, h is Plank's constant

Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J

Number of photons = Total energy / energy of one photon

N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94

6 0
3 years ago
D) 2,5 dm3 =... cm3= .... ml =...cc<br>​
DerKrebs [107]
Hey can u give a little more detail?
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2 years ago
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