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goldenfox [79]
2 years ago
6

Please answer. Thanks

Physics
1 answer:
Brilliant_brown [7]2 years ago
8 0
I believe #1. is the answer " Mitosis and specialization", because all the information needed for growth, development, and reproduction is present in a Zygote
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Which forms of energy are involved when you turn on a desk lamp and the bulb<br> becomes hot?
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Answer:

Electric energy moves through the wire to the lamp's bulb. In the bulb, electric energy transforms into thermal energy and light energy.

Explanation:

4 0
3 years ago
What is the speed of an object at a given moment
Liula [17]

Answer:

Instantaneous Speed - The speed of an object at a given moment.

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u><u>.</u></h3>
8 0
2 years ago
a balloon inflated in a room at 297k has a volume of 4.00 l. the balloon is then heated to a temperature of 331 k. what is the n
Naddika [18.5K]
V2 = 4.4579 L

Since pressure is constant, use Charle’s law.
Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

V(olume) 1 = V(olume) 2
————— —————
T(emperature) 1 T(emperature)2

4.00 L = V2
———- ———
297 K 331 K

Cross multiply
(4.00 L x 331 K) = (297 K x V2)
Simplify
1324 L/K = 297 K x V2
Isolate V2 by dividing out 297 K
1324 L/K = V2
————
297 K
(This cancels out the kelvin and leaves you with Liters as the volume measure)

V2 = 4.4579 L

Round to significant digits if required
7 0
2 years ago
6. And what's up with ChiN?
n200080 [17]
I don’t know what’s up with chin man ??
4 0
2 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
2 years ago
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