How many electrons does bromine (#35) have in its valence level?

Three joules of work is needed to shift 10 C of charge from one place to another. The potential difference between the places is

dimaraw [331]

Answer:

The potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Explanation:

Given

Work done W = 3J

Amount of Charge q = 10C

To determine

We need to determine the potential difference V between the places.

The potential difference between the two points can be determined using the formula

Potential Difference (V) = Work Done (W) / Amount of Charge (q)

or

$\:V\:=\:\frac{W}{q}$

substituting W = 3 and q = 10 in the formula

$V=\frac{3}{10}$

$V=0.3$ V

Therefore, the potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

4 0

3 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago

Unbalanced contact forces between several object would result in ?

Svetlanka [38]

In a reaction with a directing vendor of the sum forces towards some object or nothing

3 0

3 years ago

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

anzhelika [568]

To solve the problem, use Kepler's 3rd law :

T² = 4π²r³ / GM

Solved for r :

r = [GMT² / 4π²]⅓

but first covert 6.00 years to seconds :

6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s

The radius of the orbit then is :

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m

6 0

3 years ago