When pushing on a wall while wearing skates, what provides the force that causes you to accelerate away from the wall?

? Which statement is true of an object in equilibrium?

Degger [83]

The answer is C,<span> The sum of all forces acting on the object is zero. hope that helps!!</span>

7 0

3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

2 years ago

The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with

nikdorinn [45]

Answer:

A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

We are given;

x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

v = √48

v = 6.93 m/s

B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

0.015v² + 0.36 = 0.72

0.015v² = 0.72 - 0.36

v² = 0.36/0.015

v = √24

v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

v(max) = √27

v_max = 5.2 m/s

6 0

2 years ago

If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder

Lerok [7]

A baseball would hit the bat harder. This is because the baseball is a lot heavier and more dense than the plastic ball. The keyword that you're looking for is density. The baseball is dense.

7 0

3 years ago

A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the

yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0

3 years ago