Can someone help?? 3/5

An egg is thrown upward with a velocity of 4.5 m/s. How long will it take to reach it's maximum height?

Ray Of Light [21]

Answer:

0.45 seconds

Explanation:

Letting the value of g = 10 m/s/s

final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)

initial velocity(u) = 4.5 m/s

acceleration = -10 m/s/s (since the gravity is acting against the egg)

time = t seconds

From the first equation of motion:

v = u + at

0 = 4.5 + (-10)t

t = -4.5 / -10

t = 0.45 seconds

3 0

3 years ago

Why is the following situation impossible? The object of mass m = 4.00 kg in Figure P6.10 is attached to a vertical rod by two s

Crank

The situation is impossible mainly because we can't see Figure P6.10 .
It would undoubtedly be the same story on an another planet, until we
see the figure and understand what's going on.

4 0

3 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

Given:

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

Assume:

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

4 0

3 years ago

Calculate the frequency of yellow light with a wavelength of 580x10 to the -9

aniked [119]

$5.2x10¹⁴ hz$ is your answer.

3 0

2 years ago

What is the force on a 1,000 kilogram-elevator that is falling freely under the acceleration of gravity only?

likoan [24]

   Force = (mass) · (acceleration)

   = (1,000 kg) · (9.8 m/s²)

   = 9,800 newtons

Why are you still having a problem with F = M · a ?

8 0

3 years ago