The football player throws the football with an initial diagonal velocity of 17 m/s at an

When an 80.0-kg man stands on a pogo stick, the spring is compressed 0.120 m. (a) What is the force constant of the spring? (b)

Nimfa-mama [501]

Answer:

a.6533.333 N/m

b.Yes,the spring will be compressed more when the person hops down the road .

Explanation:

Mass of man=80 kg

Spring compressed by x=0.12 m

a.We have to find the force constant of the spring.

Spring constant= $k=\frac{F}{x}=\frac{mg}{x}$

$g=9.8m/s^2$

Using the formula

Spring constant, $k=\frac{80\times 9.8}{0.12}$ N/m

Spring constant= $k=6533.33 N/s$

b.The spring will be compressed more when the person hops down the road because when the person hops down then the kinetic energy of person will be converted into potential energy during the road hit.

7 0

3 years ago

Find the area of the surface.the part of the plane x 2y 3z = 1 that lies inside the cylinder x2 y2 = 6

AlladinOne [14]

I assume you mean the plane $x+2y+3z=1$ . Its area over the region

$R = \left\{(x,y) ~:~ x^2 + y^2 \le 6\right\}$

is given by the integral

$\displaystyle \iint_R dA = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy$

where $z=f(x,y) = \frac{1 - x - 2y}3$ .

We have

$\dfrac{\partial f}{\partial x} = -\dfrac13$

$\dfrac{\partial f}{\partial y} = -\dfrac23$

so that the area element is

$dA = \sqrt{1 + \left(-\dfrac13\right)^2 + \left(-\dfrac23\right)^2} \,dx\,dy = \dfrac{\sqrt{14}}3\,dx\,dy$

Then we have

$\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \iint_R dx \, dy$

and the remaining integral is exactly the area of the disk $x^2+y^2\le6$ . Its radius is √6, so its area is π (√6)² = 6π. So the area of the surface is

$\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \cdot 6\pi = \boxed{2\sqrt{14}\pi}$

8 0

2 years ago

The role of learning in motivation is most obvious from the influence of

photoshop1234 [79]

Answer:

Your answer should be Incentives

Explanation:

7 0

3 years ago

Which one of the following statements is false?

emmasim [6.3K]

Answer:

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

Explanation:

Affirmations

a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics

b) True. If both give the same results and use the same quantum number (n)

c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

7 0

4 years ago

PLEASE HELPPPPP

Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (zero movement) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

graph B, and

graph D.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0

2 years ago