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2 years ago

14

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

wer to the nearest whole number.

2 answers:

Anna71 [15]2 years ago

6 0

The Gravitationa potential energy of the mass (PEG) is given by:
$U=mgh$
where
m is the mass
g is the gravitational acceleration
h is the heigth of the mass above the reference level (the ground)

In this problem, $m=250 kg$ and $h=0.5 m$ , therefore the gravitational potential energy of the mass is:
$U=mgh=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

emmainna [20.7K]2 years ago

3 0

Answer:

i can confirm it is 1,225

Explanation:

based on the website. I just did this question.

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A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards

Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

$Y = 1.6 X - 0.032x^2$

we know that maximum height occurs is given as

$x = -\frac{b}{2a}$

$y =- \frac{1.6}{2(-0.032)} = 25$

and maximum height is

$y = 1.6\times 25 - 0.032\times 25^2$

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0

2 years ago

Read 2 more answers

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

2 years ago

A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the fi

allochka39001 [22]

Answer:

a. q2 = 16.4μC, positive charge

b. F = 0.900N

c. downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

$F_e=k\frac{q_1q_2}{r^2}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

$q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C$

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

3 0

2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

2 years ago

If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v

dexar [7]

The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units

5 0

2 years ago