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2 years ago

Superman does an exhibition run at a track meet. When he runs the 200 m

Physics

1 answer:

Ber [7]2 years ago

5 0

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered = 200m

Acceleration = 10m/s²

Unknown:

Time taken to cover the track = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

S = ut + $\frac{1}{2}$ at²

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

So;

S = $\frac{1}{2}$ at²

200 = $\frac{1}{2}$ x 10 x t²

200 = 5t²

t² = 40

t = 6.32s

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Can anyone help me? (physics)

Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

Are all liquids able to conduct an electric current

Alona [7]

The ions are able to carry electric current through a solution. Some liquids such as oil or alcohol do not form ions and do not conduct electricity. Vinegar is mostly water with a small amount of acetic acid in it. The acetic acid separates into ions on so that the solution conducts electricity. BRAINLIST ME PEASE

6 0

2 years ago

You lift a large bag of flour from the floor to a 2.5m high counter, doing 400J of work in 2 seconds. How much force did you app

oksian1 [2.3K]

<h2>The man have to apply force of 160 N</h2>

Explanation:

The work done to lift the bag of weight mg through height 2.5 m is 400 J

The work done can be found by relation W = mg x h

Thus mg = $\frac{W}{h}$ = $\frac{400}{2.5}$ = 160 N