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34kurt
3 years ago
13

A 3kg object moving at 4 m/s encounters a 20 N resistive force over a duration of 0.20 seconds. The impulse experienced by this

object is _____
Physics
1 answer:
WITCHER [35]3 years ago
3 0

Answer:

the impulse experienced by this object is 4 Ns

Explanation:

Given;

mass of the object, m = 3 kg

velocity of the object, v = 4 m/s

resistive force, F = 20 N

duration of impact, t = 0.2 s

The impulse experienced by this object is calculated as follows;

J = F x t

J = 20 x 0.2

J = 4 Ns

Therefore, the impulse experienced by this object is 4 Ns

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the focal length of a simple magnifier is 8.50 cmcm . assume the magnifier to be a thin lens placed very close to the eye.
Igoryamba

When the object is at the focal point the angular magnification is 2.94.

Angular magnification:

The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.

Here we have to find the angular magnification when the object is at the focal point.

Focal length = 6.00 cm

Formula to calculate angular magnification:

Angular magnification = 25/f

                                            = 25/ 8.5

                                             = 2.94

Therefore the angular magnification of this thin lens is 2.94

To know more about angular magnification refer:: brainly.com/question/28325488

#SPJ4

5 0
1 year ago
What is a gravitational wave and why was it so hard to detect?
Alika [10]

Answer:

gravitational waves are ripples in spece-time caused primarily when objects are accelerated and the energy for the acceleration is transpoted as gravitational radiation.

they are difficult to detect because they require very sensitive technology or you will have to wait unitl black holes collide.

5 0
3 years ago
On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s

\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m

From Hooke's law

mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0
2 years ago
An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar
Iteru [2.4K]
The direction of the electric field would be south. 

qE/m = 115 
<span>       E = 115*m/q </span>
<span>           = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span>           = 762.87 * 10^(-12) </span>
<span>           = 6.27 x 10^-10 N/C
</span>
Hope this answers the question. Have a nice day. Feel free to ask more questions.
6 0
2 years ago
15. यदि दिइएको केन्द्र 0 भएको वृत्तको
miv72 [106K]

Answer:

what language is that

Explanation:

i don't understand the languge u used please can you change it

8 0
2 years ago
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