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2 years ago

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A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t

he test charge? (k = 9.0 × 109 newton·meter2/coulomb2)

2 answers:

Novosadov [1.4K]2 years ago

8 0

Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C

Sveta_85 [38]2 years ago

8 0

Answer:

E. 9.2 × 108 newtons/coulomb.

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How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

2 years ago

A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?

murzikaleks [220]

The answer is: " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V ; That is, "mass divided by volume" ;

Density is expressed as:
__________________________________________
   "mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
_____________________________________________
{Note the exact equivalent: 1 cm³ = 1 mL }.
____________________________________________
→ The formula is: " D = m / V " ;
___________________________________________
in which:

"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);

"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;

"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
_________________________________________________
We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________
D = m / V ;
_________________________________________________________
And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
___________________________________________________
Multiply each side of the equation by "V" ;
____________________________________________________
V * { D = m / V } ; to get:
____________________________________________________
V * D = m ;   ↔ m = V * D ;
___________________________________________________
Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
______________________________________________________
m = V * D ;

m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;

→ Round to "208 g" (3 significant figures);
____________________________________
The answer is: " 208 g " .
_____________________________________________________

7 0

2 years ago

A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across f

valina [46]

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, $u_{x} = 10 m/s$

The distance between the buildings, $d_{x} = 2.0 m$

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

$h = ut + \frac{1}{2}gt^{2}$

$h = \frac{1}{2}gt^{2}$

$2.5 = \frac{1}{2}\times 10\times t^{2}$

t = 0.707 s

Now,

When the policeman was chasing across:

$d_{x} = u_{x}t + \frac{1}{2}gt^{2}$

$d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m$

The distance they will meet at:

9.57 - 2.0 = 7.57 m

8 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago

What is the sequence of energy transformations associated with a hydroelectric dam?

Hatshy [7]

Gravitational potential energy -> Kinetic energy -> Mechanical energy -> Electrical energy.

The water starts up (potential) and flows down (kinetic), the flowing water turns a big wheel (mechanical) which creates electricity (electrical).

3 0

2 years ago

Read 2 more answers