All categories

3 years ago

1. The maximum running speed (S) in km/hr and the corresponding body mass (m) in kg were

Physics

1 answer:

Ede4ka [16]3 years ago

4 0

Answer:

(a) [km/hr/kg]

(b) S[m/s] = (5.7 / 3600) (m[g])

Explanation:

(a) S[km/hr] = 5.7 (m[kg])

Solving for 5.7:

5.7 = S[km/hr] / (m[kg])

5.7 = S/m [km/hr/kg]

(b) S[km/hr] = 5.7 (m[kg])

Convert km/hr to m/s.

S[m/s] = S[km/hr] × [1000 m/km] × [1 hr / 3600 s]

S[m/s] = S[km/hr] / 3.6

3.6 S[m/s] = S[km/hr]

Convert kg to g.

m[g] = m[kg] × [1000 g/kg]

m[g] = 1000 m[kg]

m[g] / 1000 = m[kg]

Substitute.

3.6 S[m/s] = 5.7 (m[g] / 1000)

S[m/s] = (5.7 / 3600) (m[g])

You might be interested in

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$

Where,

$I_ {0}$ indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

$\theta$ indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

$I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)$

Replacing the values,

$I=100*cos^2(20)*cos^2(40-20)$

$I=100*cos^4(20)$

$I=77.91W/m^2$

Therefore the intesity of the light after it has passes through both polarizers is $77.91W/m^2$

7 0

3 years ago

What is Newton's 2nd law of motion? *<br> Your answer

Novay_Z [31]

Acceleration of an object is depended upon the net force acting open the object and the mass of the object

4 0

2 years ago

Read 2 more answers

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago

You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H

Karo-lina-s [1.5K]

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end $x=$ $3m$ , frequency of the wave $f$ = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ $v = f$ λ

Where $v =$ speed of the standing wave.

also, ∴ $v=\frac{x}{t}$

where $t =$ time take to fill entire length of the string.

Compare above both equation,

⇒ $t =$ $\frac{3}{2.3}$ sec

$t = 1.3sec$

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

7 0

2 years ago

A 1000 kg elevator is rising and its speed is increasing with an acceleration of 3 m/s^2. What is the resulting tension in the v

Alchen [17]

Answer:

12800 N

Explanation:

7 0

2 years ago