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3 years ago

Alex is driving at constant velocity of 30 mph. How far does he travel in 2.5 hours?

Physics

2 answers:

malfutka [58]3 years ago

7 0

Answer:

${\boxed {\boxed{ \sf 75 \ miles}}$

Explanation:

Distance is the product of speed and time.

$d=s*t$

The speed is 30 miles per hour and the time is 2.5 hours.

$s= 30 \ mi /hr \\t= 2.5 \ hr$

Substitute the values into the formula.

$d= 30 \ mi/hr * 2.5 \ hr$

Multiply and the hours will cancel.

$d= 75 \ mi$

The distance traveled in 2.5 hours is 75 miles.

Mandarinka [93]3 years ago

3 0

Answer: 75 miles

Explanation: Since he is driving 30 mph. You multiply how many hours he traveled. 30 x 2 hours = 60. Since he also drove half an hour you divide 30 by 2 which equals 15. So 60 + 15 = 75.

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What is significant about the primary colors of pigments?

ASHA 777 [7]

A. They can be mixed together to make almost any other color.

Explanation:

The most significant thing about primary colors is that they can be mixed together to form any other colors. Primary colors are red, blue and green. These are the three primary colors of light.

Other colors can derived from primary colors when making pigments.
If the the three primary colors are mixed together, white color is produced.
Secondary colors are the other colors produced by combining any two primary colors of light.
Two colors the produce white when mixed are complementary colors.

Learn more:

Color vision brainly.com/question/5661389

#learnwithBrainly

8 0

3 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago

A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi

Rus_ich [418]

A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + PEi
1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s

b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?

1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m

3 0

3 years ago

What's the minimum Out PUT WORK required to raise 14,0m3 of water 26.0m?

BartSMP [9]

Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

3 0

3 years ago