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3 years ago

How many miles will be completed by a runner with a constant speed of 5 mph in 30 minutes?

1 answer:

5 0

Distance=speed times time
speed=5mph=5 $\frac{mi}{hr}$
time=30mins=30/60hr=1/2hr

so
distance=speed times time
$distance=(5\frac{mi}{hr})(\frac{1}{2}hr)$
$distance=\frac{5mi \space\ hr}{2hr}$
$distance=2.5mi$

he will run 2.5 miles in 30 mins at the speed of 5mph

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LekaFEV [45]

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3 years ago

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

4 0

3 years ago

What velocity in km/hr would it take to accelerate a 2 kg object to the same momentum of a 1500 kg object with a velocity of 1.6

Lynna [10]

Answer:

v = 4374 Km/h

Explanation:

Given that,

Mass of the smaller object, m = 2 Kg

Mass of the bigger object, M = 1500 Kg

Velocity of the bigger object, V = 1.62 m/s

Velocity of the smaller object, v = ?

The product of its mass and velocity of a body is equal to its linear momentum. It is given by the formula

p = mv Kg m/s

To find the momentum of the bigger object, substitute M and V in the above equation

p = 1500 Kg x 1.62 m/s

= 2430 Kg m/s

The velocity imparted to the small body to attain this momentum is given by the relation

v = p/m m/s

= 2430 Kg m/s / 2 Kg

= 1215 m/s

By converting the velocity to Km/h

v = 4374 km/h

Hence, the velocity of the 2 Kg object is v = 4374 km/h

6 0

3 years ago

A small aircraft accelerated down a runway at 4.0 m/s²

Radda [10]

Given data in the problem :-

Acceleration (a) = 4.0 m/s^2
Initial velocity (u) = 0 m/s
Final velocity (v) = 34 m/s
Distance travelled by aircraft (S) = ?

From Newton's Laws of Motion we know that ,

v = u + at [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴ t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴ S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .

4 0

2 years ago

A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicul

Fantom [35]

QUESTION:

Part A

The induced emf in the loop is measured to be $V$ . What is the magnitude $B$ of the magnetic field that the loop was in?

Part B

For the case of a square loop of side length $L$ being pulled out of the magnetic field with constant speed $v$ (see the figure), what is the rate of change of area $c = -\dfrac{dA}{dt}$ ?