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lubasha [3.4K]
3 years ago
14

What is the voltage across a load that has a resistance of 7.5 ohms and a 1.8 A current flowing through it? *

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

13.5V

Explanation:

u=i*r

u=1.8×7.5=13.5V

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3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
3 years ago
Two molecules of lithium are combined with 225 grams of bromine to form two molecules of lithium bromide. If you end up with 690
4vir4ik [10]

The amount of Li present to start the reaction is 55.18g

<u>Explanation:</u>

2Li + Br₂ → 2LiBr

Molecular weight of Br₂ = 159.808 g/mol

Mass of Br₂ present = 225 g

Moles of Br₂ present during the reaction = 225 / 159.808

                                                                m = 1.4

Molecular weight of LiBr = 86.845 g/mol

Mass of LiBr formed = 690 g

moles of LiBr produced = 690 / 86.845

                               m(LiBr) = 7.95

According to the balanced equation, 2 molecules of Li reacts to for 2 molecule of LiBr

So, 7.95 moles of LiBr would require 7.95 moles of Li

The molecular weight of Li is 6.941 g/mol

Thus, the amount of lithium present to start the reaction is

moles = \frac{given weight}{molecular weight} \\\\7.95 = \frac{w}{6.941} \\\\w = 55.18g

Therefore, the amount of Li present to start the reaction is 55.18g

6 0
3 years ago
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