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BabaBlast [244]
3 years ago
15

5 grams is thes than greater than or equal too 508mg

Physics
2 answers:
guapka [62]3 years ago
7 0

Answer:

yes 5 is greater

Explanation:

makkiz [27]3 years ago
4 0

Answer: 5 grams is greater than 508 milligrams. Even though the number '508' is larger than '5,' you need to look at the units.

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A train has an initial velocity of 44m/s and an accelaration of _4m/s calculate its velocity​
Kobotan [32]

Complete question:

A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity​ after 10s ?

Answer:

the final velocity of the train is 4 m/s.

Explanation:

Given;

initial velocity of the train, u = 44 m/s

acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)

time of motion, t = 10 s

let the final velocity of the train = v

The final velocity of the train is calculated using the following kinematic equation;

v = u + at

v = 44 + (-4 x 10)

v = 44 - 40

v = 4 m/s

Therefore, the final velocity of the train is 4 m/s.

7 0
3 years ago
Convert 4 µL to mL<br> Please help meeeee
stiv31 [10]

Answer: the answer would be four thousand

Explanation: hope this helps

3 0
3 years ago
Read 2 more answers
PLEASE I NEED THIS NOW ITS EASY I JUST DIDNT STUDY
dem82 [27]

Answer:

Explanation:

Frequency*speed

= 0.5*2.5

= 1.25m

3 0
2 years ago
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of
Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

E = \frac{mg}{q}

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

q = \frac{mg}{E}

Replacing,

q = \frac{(3.37*10^{-9})(9.8)}{11000}

q = 3.002*10^{-12}C

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

q = ne

Here,

n = Number of electrons

e = Charge of each electron

n = \frac{q}{e}

Replacing,

n = \frac{3.002*10^{-12}}{1.6*10^{-19}}

n = 2.44*10^7

Therefore the number of electrons that reside on the drop is 2.44*10^7

5 0
3 years ago
A coin is dropped in a 15.0 m deep well.
labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0
2 years ago
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