All categories

3 years ago

Sweating and panting are examples of which characteristics of life?

Physics

2 answers:

Marianna [84]3 years ago

7 0

C.reproducing is think and if wrong I'm so very sorry

Debora [2.8K]3 years ago

3 0

B. using energy. using alot of energy, like for exercising causes sweating and panting

You might be interested in

Interactive LearningWare 8.1 reviews the approach that is necessary for solving problems such as this one. A motorcyclist is tra

STALIN [3.7K]

Answer:

The angular displacement of each wheel is 269.92 rad

Explanation:

Given:

Angular acceleration $\alpha = 6.10 \frac{rad}{s^{2} }$

Time to pass cyclist $t = 4.36$ s

Angular velocity $\omega _{f} = 75.2 \frac{rad}{s}$

According to the equation of kinematics,

$\omega _{f} = \omega _{i} + \alpha t$

$\omega _{i} = \omega _{f} - \alpha t$

$\omega _{i} = 75.2 - 6.10 \times 4.36$

$\omega _{f} = 48.60$ $\frac{rad}{s}$

For finding angular displacement,

$\omega _{f} ^{2} - \omega _{i} ^{2} = 2 \alpha \theta$

Where $\theta =$ angular displacement,

$\theta = \frac{\omega _{f}^{2} - \omega _{i} ^{2} }{2\alpha }$

$\theta = \frac{5655.04 - 2361.96 }{2\times 6.10 }$

$\theta = 269.92$ rad

Therefore, the angular displacement of each wheel is 269.92 rad

8 0

3 years ago

A point charge q1 = -5.7 μC is located at the center of a thick conducting spherical shell of inner radius a = 7.4 cm and outer

olga55 [171]

Answer: a) see attach file; b) E=0; c)E=-1.65* 10^6 N/C

Explanation: In order to solve this problem we have to use the gaussian law which is given by:

∫E*dr=Q inside/εo

E for r=8 cm is located inside the conducting shell so E=0.

For r=13 cm we have to use above gaussian law considering the total charge inside a gaussian surface with radius equal to 0.13 m. ( see attach)

4 0

3 years ago

What is the value of Vo in the

notsponge [240]

Answer: 60m/s

Explanation:

From the diagram:

Θ = 30°

Vertical resolution (y-axis) :

Voy = VoSinΘ

g in the upward direction = negative (-) = - g

Vfinal = 0

Distance (H) traveled along y =

Time taken to reach maximum height :

From v = u + at

0 = usinΘ - gt

gt = usinΘ

t = usinΘ / g

Horizontal resolution:

S = ut + 1/2at^2

Substituting t = usinΘ / g ; Voy = usinΘ

S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2

S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)

S = (u^2sin^2Θ) / 2g

Now if S = maximum height = 45m

Then,

45 = [Vo^2sin^2(30°)] / 2(10)

45 =[ Vo^2 * (0.5)^2] / 20

45 =( Vo^2 * 0.25) / 20

20 * 45 = Vo^2 * 0.25

900 / 0.25 = Vo^2

3600 = Vo^2

Vo = sqrt(3600)

Vo = 60m/s

5 0

3 years ago

In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a

katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

brainly.com/question/11422992

#SPJ1

3 0

2 years ago

Please help me with question B.

Step2247 [10]

It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).

5 0

3 years ago