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Vadim26 [7]
3 years ago
8

If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no l

onger any force on the ball due to the woman's hand), what is the relationship between aaa and ggg, the magnitude of the acceleration of gravity
Physics
1 answer:
kakasveta [241]3 years ago
5 0

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

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1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es
bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which  the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h

or

v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s

So, the required speed is 15 km/h or 4.16 m/s.

7 0
3 years ago
MATHPHYS HELP
rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0
3 years ago
At what distance from the centre of the earth its acceleration due to gravity becomes one third only? [Mass and radius of earth
Kipish [7]

Answer:

235

Explanation:

3 0
3 years ago
Read 2 more answers
Scientific observations should be reported with bias as long as they benefit a scientist's employer
Ira Lisetskai [31]
False, there are always pros and cons to research
8 0
3 years ago
A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential
BARSIC [14]

Answer:

1.88\cdot 10^{-5} A

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{\epsilon_0 A}{d} (1)

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

Q=CV (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

J=\frac{Q}{t} (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}

where we have

A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2

d = 0.58 mm = 5.8\cdot 10^{-4} m

\frac{V}{t}=500,000 V/s

Substituting into the equation, we find

J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A

6 0
3 years ago
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