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baherus [9]
3 years ago
10

The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the drive

r's reaction time of 0.50 s . What is the minimum stopping distance for the same car traveling at a speed of 50
Physics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

150 m

Explanation:

Distance traveled during reaction time = .5 x 30 = 15 m

Actual distance during which there was deceleration = 60 - 15 = 45 m .

force x displacement = kinetic energy

F x s = 1/2 m v² ;

F x 45 = 1/2 m 30² = 450 m ----( 1 )

for second case

Fx S = 1/2 m 50² = 1250 m --------( 2 )

Dividing( 1 )and ( 2 )

S/45 = 1250 /450 = 125 / 45

S = 125 m

Add distance traveled in reaction time in second case = 50 x .5 = 25 m

total distance required = 25 + 125 = 150 m

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On an extremely warm day, the balloon might pop because gases expand the hotter they get, and due to its temperature it is likely to pop if it is, indeed, nearly, if not completely, filled to its capacity.  I hope this helps, have a nice day!
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Which of the following would have the most momentum?
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I think it should be D as momentum is the product of mass and velocity...
4 0
2 years ago
Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F> 32° C>32 K 32°C 32° F 32 K 32° F 32 K 32° c
Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

3 0
3 years ago
A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13
tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

F = qvB

here we have

B = \frac{F}{qv}

here we know that

F = 4.8 \times 10^{-13} N

q = 1.6 \times 10^{-19} C

v = 4 \times 10^6 m/s

now from above equation we have

B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}

B = 0.75 T

8 0
3 years ago
Two metal spheres are suspended from strings. The
nikdorinn [45]

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = -2 \times 10^2 C

q_2 = -4 \times 10^{-8} C

r = 3.0 m

now we have

F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}

F = 8000 N

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

3 0
3 years ago
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