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AfilCa [17]
3 years ago
10

How did you know if an electric current is flowing in a lightbulb?

Physics
1 answer:
sweet [91]3 years ago
7 0

Answer:

if it lights up

Explanation:

electricity

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The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y
xeze [42]

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

                                    \frac{30}{2} = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

8 0
3 years ago
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 22.0 m/s and returns the shot with the ball
morpeh [17]
I need an answer choice
8 0
3 years ago
A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with
padilas [110]
Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s

A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s

B) momentum of the raft = - momentum of the girl = -24.0 N/s

C) speed of the raft

p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]
8 0
3 years ago
Enzyme reaction and deactivation Lipase is being investigated as an additive to laundry detergent for removal of stains from fab
natita [175]

Answer:

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]

Explanation:given values

Half life of lipase t_1/2 = 8 min x 60s/min = 480 s

Rate constant for first order reaction

k_d = 0.6932/480 = 1.44 x 10^-3 s-1

Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L

rate of hydrolysis Vm0 = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3

K_m = 5mmol/L

time take is given by

t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]

all values are given and putting these value we get

t=1642.83 secs

which is equal to

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]

5 0
3 years ago
ANSWER QUICKLY IN LESS THAN A MINUTE!!EASY!​
luda_lava [24]

Answer:

I think it's 0 N 3rd choice

4 0
2 years ago
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