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Using science terms like force, mass, energy, and inertia, describe how you could get an object to fly a further distance in a c

qaws [65]

i need the same question

5 0

2 years ago

A free-falling golf ball strikes the ground and exerts a force on it. Which sentences are true about this situation? A golf ball

Harlamova29_29 [7]

Answer:

The ground exerts an equal force on the golf ball

Explanation:

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

In this problem, object A is the golf ball while object B is the ground, so we can say that:

- the golf ball exerts a force on the ground

- the ground exerts an equal and opposite force on the golf ball

8 0

2 years ago

Read 2 more answers

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

a) Determine the angular accelerations of B1 and B2 when couple is applied

at point P ; Co = I* ∝B2'

150 = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 = 180 rad/sec

b) Determine the force exerted on B2 at P

T2 = mB1g + T1 -------- ( 1 )

where ; T1 = mB2g ( at point p )

= 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

3 0

3 years ago

A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping

leva [86]

Answer:

0.02896 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A21$ = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

$x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)$

At maximum displacement

$cos(\omega t+\phi)=1$

$\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s$

The magnitude of the damping coefficient is 0.02896 kg/s

6 0

3 years ago

Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an lrc-series circuit is l d 2q dt 2

Aloiza [94]

DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150

5 0

2 years ago