All categories

2 years ago

WILL GIVE BRAINLIEST

2 answers:

8 0

Mass of girl =55kg
acceleration due to gravity =9.8m/s^2

So, the force she is applying on the floor must equal
55× 9.8 N
(Because force =mass ×acceleration)
=539N

Now, according to newton's third law the floor must be applying the same force on the body in the opposite direction.
So the force applied here by the floor is 539N upward.

Option "C" is the correct one.

Rudik [331]2 years ago

7 0

The answer is at least 539 N up

You might be interested in

A particle of unit mass moves in a potential V(x)= ax^2+b/x^2 . Where a and b are constnts. Calcuate the angular frequency of sm

Sholpan [36]

Answer:

Explanation:

Given

Potential Energy is given by

$U(x)=ax^2+\frac{b}{x^2}$

And Force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

Particle will be at equilibrium when Potential Energy is either minimum or maximum

$F=-\left ( 2ax-\frac{2b}{x^3}\right )$

i.e. $ax=\frac{b}{x^3}$

$x_0=(\frac{b}{a})^{0.25}$

So angular Frequency of small oscillation is given by

$\omega =\sqrt{\frac{U''(x)}{m}}$

for $m=1$

we get $\omega =\sqrt{\frac{U''(x_0)}{1}}$

$U''(x_0)=2a+6a= 8a$

$\omega =\sqrt{8a}$

4 0

2 years ago

A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate

irina1246 [14]

Perimeter = 2 ( L + W )
32 = 2 ( L + W )
16 = L + W
L = 16 - W

Area = L W
63 = L W
63 = (16-W) W
63 = 16W - W²
-W² + 16 W - 63 = 0
By factorizing W = 9 or W = 7
So the dimensions are 7 and 9

3 0

2 years ago

An LED operation at 850 nm center wavelength has a spectral width of 45 nm. What is the pulse spreading in ns/km

Papessa [141]

Answer:

$\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

Explanation:

From the given information, the LED is operating with a given wavelength of 850 nm or 0.85 μm.

Hence, the material dispersion is $\dfrac {d \tau _{mat}}{d \lambda } \simeq (80 \ ps / (nm.km) \ )$

Now, using the pulse spread formula:

$\dfrac{\sigma_{mat}}{L} = \dfrac{d \tau _{mat} }{d \lambda} \sigma \lambda$

$\dfrac{\sigma_{mat}}{L} = (80 \ ps/ ( m.km) \ ) \times (45 \ nm)$

Thus, the pulse spreading as a result of material dispersion is: $\mathbf{\dfrac{\sigma_{mat}}{L} = 3.6 \ ns/ km}$

3 0

2 years ago

Need help with 2 and 3<br>pls help, due in 1 1/2 hours<br>(GIVING 20 POINTS)

Anarel [89]

2) acceleration = final velocity - initial velocity / time —> V-U/T
Acceleration is the change in velocity over the change in time so it can be represented by the equation a = Δv/Δt.
3) first one- F=10.5 N
second one- 4 m/s^2
third one- 1200N

7 0

3 years ago

Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th

Vladimir79 [104]

Answer:

$C = 1.77 \times 10^{-4} F$