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2 years ago

Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.

Physics

1 answer:

lilavasa [31]2 years ago

7 0

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have .21 ( is that down? I will assume it is)

Counterclockwise moments :

.21 * 40 + 1.0 * 20

Clockwise moments :

.5 * 20 + F * 45

these moments must equal each other

.21*40 + 1 *20 = .5 * 20 + F * 45

F = .409 N

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Buoyant force is the net upward force that affects on the object in a fluid

Katarina [22]

Answer:True

Explanation:

Buoyant force is the net upward force, that affect on the object in a fluid

4 0

3 years ago

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about

nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0

3 years ago

Weigt A box sitting still on the ground by itself has a weight of 700N, what is the mass? (gravity's acceleration is 9.80 m/s2)

Alex

Force=700N
Acceleration=9.8m/s^2

Using newtons law

$\\ \rm\longmapsto F=ma$

$\\ \rm\longmapsto m=\dfrac{F}{a}$

$\\ \rm\longmapsto m=\dfrac{700}{9.8}$

$\\ \rm\longmapsto m=71.4kg$

7 0

2 years ago

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

zlopas [31]

Answer:

q = 3.6 10⁵ C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

r = 6 , 37 106 m

E = k q / r²

q = E r² / k

q = $\frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}$

q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

r = 0.6 r_ Earth

r = 0.6 6.37 10⁶ = 3.822 10⁶ m

E = k q / r²

q = E r² / k

q = $\frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}$

q = 3.6 10⁵ C

4 0

3 years ago

A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.

olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

5 0

3 years ago