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grigory [225]
3 years ago
11

At what speed will a car round a 52-m-radius curve, banked at a 45???? angle, if no friction is required between the road and ti

res to prevent the car from slipping? (g = 9.8 m/s2)a. 27 m/s
b. 17 m/s
c. 23 m/s
d. 35 m/s
Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer:

c. 23 m/s

Explanation:

\theta = Angle of banking = 45°

r = Radius of turn = 52 m

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of car

As the car is tilted we have the relation

tan\theta=\dfrac{v^2}{rg}\\\Rightarrow v=\sqrt{tan\theta rg}\\\Rightarrow v=\sqrt{tan45\times 52\times 9.8}\\\Rightarrow v=22.57432\ m/s\approx 23\ m/s

The velocity of the car is 23 m/s if the car is not slipping

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Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of
Alchen [17]

Answer:

\theta=10.20^{\circ}  

\Delta l=0.10 ft    

Explanation:

First of all, we analyze the system of blocks before starting to move.

\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

16sin(\theta)-2.91cos(\theta)=0  

tan(\theta)=0.18  

\theta=arctan(0.18)  

\theta=10.20^{\circ}  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

P_{A}sin(\theta)-F_{fA}-F_{spring}=0

Where:

F_{spring} = k\Delta l=2.1\Delta l

P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

\Delta l=0.10 ft    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0
4 years ago
PLEASE ANSWERRRRR ASAPPPPPP
Bad White [126]
D Valence
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^^^answer
5 0
3 years ago
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Sherlock Holmes examines a clue by holding his magnifying glass (with
Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

M=-\frac{q}{p}=-\frac{-30}{10}=3

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0
4 years ago
Counting your pulse can be most accurately performed by applying light pressure to an artery that is close to the surface of the
pochemuha
True i believe because people use this with the wrist and the neck 
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The equation shows neutralization of an acid and a base to produce a salt and water.
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The answer is; 2.

To balance a chemical equation, the moles on one side of the equation has to be the same as that on the other side. This ensures that the law of conservation  is observed because matter or energy can't be created or destroyed but can only be transformed from one form to another.

In this equation, putting 2 in front of NaCl ensures that there are 2 moles of Na and CL just as there are 2 moles of Na and CL in the reactants side.


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3 years ago
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