Which statement is true?

HELPPP!!! During a soccer game, a player kicks the winning goal. When is work being done by the player?

zimovet [89]

Answer:

when the player’s foot is in contact with the ball

Explanation:

7 0

3 years ago

Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side suppor

TEA [102]

Answer:

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (<em>in this case due to gravity</em>) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re- arrange the equation

ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F)/(Y x L), substitute the values

ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m

6 0

4 years ago

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

4 0

3 years ago

HELP ASAP!!!! Which of the following is an example of a hypothesis? A) What will happen if a sugar cube in boiling water? B) I t

Maurinko [17]

Answer:

C) If an ice cube is placed into a boiling water, then it will melt in less than 2 minutes.

Explanation:

7 0

3 years ago

Under electrostatic conditions, the electric field just outside the surface of any charged conductor

KatRina [158]

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

3 0

3 years ago