All categories

3 years ago

Which action involves the most efficient conversion of electrical energy into thermal energy?

Physics

1 answer:

azamat3 years ago

7 0

Answer:

the answer is option d

because in oven which works through electricity ,it will bake the cake (heat energy is produced)

You might be interested in

What is analogy a powerful rhetorical device?

wolverine [178]

The other person who answered this is wrong btw

5 0

2 years ago

ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of

sesenic [268]

Answer:

a) $V_{2/1}=0.8m/s$

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

$V_{2/1} = V_{2} - V_{1} = 0.8m/s$

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

$t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s$

$t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s$

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

6 0

2 years ago

Read 2 more answers

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

Under certain circumstances, potassium ions (K+) in a cell will move across the cell membrane from the inside to the outside. Th

choli [55]

Answer:

$1.368\times 10^{-20}\ J$

Explanation:

q = Charge in the potassium ion = $19e-18e$

e = Charge of electron = $1.6\times 10^{-19}\ C$

$V_2-V_1$ = Change in potential = $0-(-85.5\times 10^{-3})$

Change in electric potential is given by

$E=q(V_2-V_1)\\\Rightarrow E=(19e-18e)(0-(-85.5\times 10^{-3})\\\Rightarrow E=1.6\times 10^{-19}\times 85.5\times 10^{-3}\\\Rightarrow E=1.368\times 10^{-20}\ J$

The energy is $1.368\times 10^{-20}\ J$

3 0

3 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

iris [78.8K]

Answer:

Explanation:

1. $V_{x}$ = $V_{0}$ * cos $\alpha$ ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. $V_{y}$ = $V_{0}$ * sin $\alpha$ ⇒ 16* sin32 ≈ 9.4 m/s

3. $y_{max}$ = $\frac{v_{0}^2*sin^2\alpha}{2g}$ = $\frac{16^2*sin^232}{2*9.8}$ (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

$y_{max}$ ≈ 3.6677+1.5 ≈ 5.2m

4. $x_{max}$ = $\frac{v_{0}^2*sin(2\alpha)}{g}$ = $\frac{16^2*sin(2*32)}{9.8}$ ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

3 0

2 years ago