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Murrr4er [49]
3 years ago
9

Name the force between the brakes and the wheels that does work to slow down the vehicle when the brakes are applied. ​

Physics
1 answer:
suter [353]3 years ago
3 0

Answer:

Friction

Explanation:

Because the brake pad is rubbing alongside the wheel to slow the body of the car down

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Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o
natulia [17]

Answer

given,

Speed of car A = 95 Km/h

                         = 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

                         = 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

   d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t  = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

   D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km  

8 0
3 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
Which of the following describes the products of a chemical reaction? A. The original materials B. The substances that are chang
lianna [129]

The products of a chemical reaction are the substances that are changed and the chemicals on the right side of a chemical equation. The correct options are B and C.

<h3>What is chemical reaction?</h3>

The chemical reaction  is the reaction between two reactants which led to the formation of products.

The products are substances which forms after reaction. The reactants are the substances which are original materials.

The reactants lie on the left side and products lie on the right side of the reaction.

Thus, the correct options are B and C.

Learn more about chemical reaction.

brainly.com/question/22817140

#SPJ1

7 0
2 years ago
One of the signs a chemical reaction is taking place
ladessa [460]

Answer:  One of the signs a chemical reaction is taking place is the presence of bubbles. :)

6 0
3 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
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