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hichkok12 [17]
3 years ago
9

If a bike rider travels 4 km in an hour,what is his speed measured in miles per hour?

Physics
2 answers:
Ugo [173]3 years ago
8 0

Answer:

2.485 miles/hour

Explanation:

Speed of the rider  is 4 km/hour

The conversion factor is

1 km = 0.621371 miles

4 \frac{km}{hour} = 4 \frac{km}{hour} \times \frac{0.621371 miles}{1 km} \\\\=2.485484 \frac{miles}{hour} \\

tatyana61 [14]3 years ago
3 0
<span>2.4854847</span> miles per hour
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Which of the following statements below is correct?
Mamont248 [21]
C. is correct. when you make a pizza, you see the the meat or pepperoni or cheese heating up and sometimes melted. (thats physical). on the inside the crust in heated and the toppings are cooked (chemical)
5 0
3 years ago
Guys please helpp!!!!1
Setler79 [48]

Answer:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy (E) is the sum of gravitational potential energy (U) and translational kinetic energy (K), all measured in joules. In addition, gravitational potential energy is directly proportional to height (h) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

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3 0
2 years ago
Part A Determine the magnitude of the x component of F using scalar notation. Fx F x = nothing lb Request Answer Part B Determin
maksim [4K]

As we know that force F makes an angle of 60 degree with X axis

so the X component is given as

cos60 = \frac{F_x}{F}

now we have

F_x = F cos60

F_x = 0.50 F

Similarly we know that force F makes an angle of 45 degree with Y axis

so the X component is given as

cos45 = \frac{F_y}{F}

now we have

F_y = F cos45

F_y = 0.707 F

Now for the component along z axis we know that

F_x^2 + F_y^2 + F_z^2 = F^2

now plug in all components

(0.707 F)^2 + (0.50 F)^2 + F_z^2 = F^2

0.5 F^2 + 0.25 F^2 + F_z^2 = F^2

F_z^2 = F^2(1 - 0.75)

F_z^2 = 0.25 F^2

F_z = 0.5 F

5 0
3 years ago
How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J
algol13

Answer:

Work done by the frictional force is 3.41\times 10^5\ J

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W=k_f-k_i

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)

W = −340605 J

or

W=3.41\times 10^5\ J

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6 0
3 years ago
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
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3 years ago
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