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Neporo4naja [7]
3 years ago
14

Please select the word from the list that best fits the definition Used to compare two specific variables

Physics
2 answers:
Ann [662]3 years ago
7 0

i would say line graph


mr Goodwill [35]3 years ago
5 0

Answer: a. line graph

Explanation:   A line graph is formed by joining the points given by the data with straight lines.

A line graph is usually used to show the change of information over a period of time using two variables . This means that the horizontal axis is usually a time scale, for example minutes, days, months or years; and the vertical axis shows the value of the information that varies in time.

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According to Ohm’s law, which combination of units is the same as the unit for resistance? volt ÷ ampere ampere × volt volt + am
Maksim231197 [3]

Answer:

volt ÷ ampere

Explanation:

The mathematical form of Ohms law is given by :

V = IR

Where V is voltage

I is current

R is resistance

R=\dfrac{V}{I}

The unit of voltage is volt and that of current is ampere

Unit of resistance :

R=\dfrac{\text{volt}}{\text{ampere}}

So, volt ÷ ampere is the same as the unit of resistance. Hence, the correct option is (a).

6 0
2 years ago
Read 2 more answers
If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take
kifflom [539]

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

8 0
3 years ago
A 15 kg dog jumps out a stationary sled which has a mass of 40 kg. If
zzz [600]
Velocity of the sled is 3.2 m/s
5 0
2 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
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