Ask question

Ask question

All categories

3 years ago

15

Help plssssssssssssssssssss

2 answers:

marta [7]3 years ago

8 0

Yes, that is most like the answer but I’m not to sure

a_sh-v [17]3 years ago

3 0

Answer:

"humans cannot hear all the same sound frequencies as animals", & "Humans cannot make make all the same sound frequencies as animals" are correct. the other two are incorrect, hope this helps!

You might be interested in

Which of the following is an example of resourcefulness?

tatiyna

A. Discovering a quick way to handle a new problem

5 0

3 years ago

What happens when the temperature of an object decreases?

Alja [10]

Answer is Option C

hope it helps you

4 0

3 years ago

Read 2 more answers

A car goes from rest to a velocity of 108 km/h north in 10s what is the car's acceleration in m/s2

fomenos

initial velocity of the car given as

$v_i = 0$

final velocity is given as

$v_f = 108 km/h$

as we know that

$1 km/h = 0.277 m/s$

now we can convert final speed into m/s

$v_f = 108 * 0.277 = 30 m/s$

now acceleration is rate of change in velocity

$a = \frac{v_f - v_i}{t}$

$a = \frac{30 - 0}{10}$

$a = 3 m/s^2$

so the acceleration of the car is 3 m/s^2

7 0

3 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

3 years ago

In a reverse fault, the fault part that lies below the other part is called the _____.

jeyben [28]

Answer:

D

Explanation:

7 0

3 years ago