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2 years ago

If a light ray from a candle strikes a mirror at an angle of 40° with the normal, at what angle will the ray reflect?

Physics

1 answer:

pav-90 [236]2 years ago

8 0

Answer:

the correct answer is b: 25

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If two materials are made from different kinds of atoms, what else should you expect to be different?

Volgvan

<span>B. The properties they have</span>

3 0

3 years ago

Read 2 more answers

An object in static equilibrium has a coefficient of static friction as 0.110. If the normal force acting on the object is 95.0

igomit [66]

Answer:

mass = 9.7 kg

Explanation:

As we know that when object is at rest on the ground of flat base then we will have

$N - mg = 0$

so from here

$N = mg$

now we have

N = 95 Newton

now from above equation we will have

$95 = m\times g$

$95 = m\times 9.8$

$m = 9.7 kg$

7 0

3 years ago

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago

Two persons manage to push a motorcar of mass 1200 kg at uniform velocity along a level road, The same motorcar can be pushed by

densk [106]

Answer:

40N by each person.

Explanation:

8 0

2 years ago

Q = cmAT

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

8 0

2 years ago