Ask question

Ask question

All categories

Romashka-Z-Leto [24]

2 years ago

14

In 1994, Leroy Burrell of the United States set what was then a new world record for the men’s 100 m run. He ran the 1.00  102

m distance in 9.5 s. Assuming that he ran with a constant speed equal to his average speed, and his kinetic energy was 3.40  103 J, what was Burrell’s mass?

1 answer:

vovikov84 [41]2 years ago

5 0

Answer:

61.33 Kg

Explanation:

From the question given above, the following data were obtained:

Distance = 1×10² m

Time = 9.5 s

Kinetic energy (KE) = 3.40×10³ J

Mass (m) =?

Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:

Distance = 1×10² m

Time = 9.5 s

Velocity =?

Velocity = Distance / time

Velocity = 1×10² / 9.5

Velocity = 10.53 m/s

Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:

Kinetic energy (KE) = 3.40×10³ J

Velocity (v) = 10.53 m/s

Mass (m) =?

KE = ½mv²

3.40×10³ = ½ × m × 10.53²

3.40×10³ = ½ × m × 110.8809

3.40×10³ = m × 55.44045

Divide both side by 55.44045

m = 3.40×10³ / 55.44045

m = 61.33 Kg

Thus, the mass of Leroy Burrell is 61.33 Kg

You might be interested in

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

$P = mv$

$P = 1.25 \times 84.3$

$P = 105.4 kg m/s$

6 0

2 years ago

A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is

Sonja [21]

Answer:

0.8895m

Explanation:

Cable diameter = 0.0125m

Mass of elevator = 6450kg

Young Modulus(E) = 2.11*10¹¹N/m

∇l (change in length) =

L = 362m

A = Πr², but r = d / 2 = 0.0125 / 2 = 0.00625m

A = 3.142 * (0.00625)² = 1.227*10^-4m²

Young Modulus (E) = Tensile stress / Tensile strain

E = (F / A) / ∇l / L

F = mg = 6450 * 9.8 = 63210N

2.11*10¹¹ = (63210 / 1.22*10^-4) / (∇l / 362)

2.11*10¹¹ = 5.18*10⁸ / (∇l / 362)

2.11*10¹¹ = (5.18*10⁸ * 362) / ∇l

2.11*10¹¹ = 1.875*10¹¹ / ∇l

∇l = 1.875*10¹¹ / 2.11*10¹¹

∇l = 0.8895m

The change in length is 0.8895m

8 0

2 years ago

A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to

nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket $V=10\ gallon$

Time taken to fill the bucket $t=50\ s$

so volume flow rate is $\dot{V}=\frac{10}{50}=0.2\ gal/s$

1 gal is equivalent to $0.133\ ft^3$

$\dot{V}=0.0267\ ft^3/s$

mass flow rate $\dot{m}=\rho \times \dot{V}$

$\dot{m}=62.4\times 0.0267$

$\dot{m}=1.668\ lbs$

(b)Average velocity through nozzle exit

$\dot{V}=Av_{avg}$

$v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}$

$v_{avg}=49.51\ ft/s$

8 0

3 years ago

________ reaction time involves selecting a specific and correct response from several choices when presented with several diffe

lukranit [14]

Simple reaction time involves selecting a specific and correct response from several choices when presented with several different stimuli. This is very important because historically, this was the first indicators of intelligence pioneered by Francis Galton. To measure one’s intelligence is to know how he quick a person could respond to the stimulus with an already expected response wherein the stimulus is given unknown to the receiver. In other terms, the intelligence is measured on how quick a person could grasp certain concepts and how he could think fast and answer them correctly. <span> </span>

4 0

2 years ago

Read 2 more answers

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0

2 years ago