We dont see objects. We see the light ___

Pls answer the 15a answer i cant understand it

Maksim231197 [3]

Materials required for the experiment of limiting force borne by string:-

String balance
weights
light strings
weight hanger
pan for spring balance
Sand

Steps of procedure for for the experiment of limiting force borne by string:-

First we have to tie a light string to the fixed support and then tie the other end with the weight hanger consists of weight.
Add additional weight to the hanger again and again. And continue the same until the string is broken.
Note down the weight (x) where the string is broken.
Suspend spring balance to a support.
Tie the light string at the end of the balance and at the other end suspend the pan for spring balance.
Now place the weights (x-100 grams) in pan.
Observe the reading in the spring balance.
Add a small amount of sand in the pan by observing the readings.
same is to be done till the string is broken.

Learn more about limiting force here:- brainly.com/question/11371672

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6 0

1 year ago

PLEASE HELP! 2ND TIME I ASKED THIS!!!

Goshia [24]

The train is moving at 50 m/s and Emma is walking down the aisle with 1 m/s speed in the same direction of train. The relative velocity of Emma with respect to other passengers pf the train would be 1 m/s. This is because, the train is not moving relative to them and only emma is moving at 1 m/s. If a person observes from outside, Emma would have (50 +1) m/s = 51 m/s velocity.

relative velocity when two objects are moving in same direction as oberved from outside observer:

$v_{ab}=v_a+v_b$

relative velocity when two objects are moving in the opposite directionas oberved from outside observer:

$v_{ab}=v_a-v_b$

4 0

3 years ago

1) How is the temperature of a gas related to the kinetic energy of its particles?

dlinn [17]

The average kinetic energy of a gas particle is directly proportional to thetemperature. An increase intemperature increases the speed in which the gas molecules move. Allgases at a given temperature have the same average kinetic energy. Lightergas molecules move faster than heavier molecules.

6 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(c) v = 0.31m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

3 years ago

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass

devlian [24]

Answer:

The velocity of the recoil is $v=1.001 \frac{m}{s}$

Explanation:

Kinetic Energy

$m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\m_{gun}= 15.0kg+3.6kg$

The mass of th gun is the both mass the shotgun and the arm shoulder combination

$m_{bullet}=0.049kg\\v_{bullet}=380\frac{m}{s} \\m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\0.049kg*380\frac{m}{s}=(15.kg+3.6kg)* v_{recoil}\\v_{recoil}=-\frac{18.62 kg \frac{m}{s} }{18.6 kg}\\ v_{recoil}=-1.0010 \frac{m}{s}$

The velocity is negative because is in opposite direction of the bullet

6 0

2 years ago

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We dont see objects. We see the light ____ off objects.