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3 years ago

A 1.5 kg rock is dropped from a height of 2.0 meters onto a spring that

Physics

1 answer:

Maslowich3 years ago

7 0

Answer:

29.4 J

Explanation:

Before the drop, the system has only the gravitational potential energy, and this energy us given by mass×gravity×height:

1.5•9.8•2 = 29.4 J

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The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive

77julia77 [94]

Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive

4 0

3 years ago

I need help with...

Rom4ik [11]

The car will gain new momentum if it's velocity is doubled or tripled.

6 0

4 years ago

Ann walked 1.5 miles south to her house in 0.5 hours. what is Ann's speed? what is Ann's velocity

ipn [44]

To calculate the speed and velocity of the Ann`s we use the formula,

$v= \frac{d}{t}$

Here, d is distance and t is time and v if we take it with direction then it is called velocity and if we take it without the direction then it is called speed.

Given $d=1.5 miles$ and $t=0.5 hours$ .

Substituting these values in above equation we get

$v=\frac{1.5mi }{0.5 h} = 3 mi/h$

As Ann walked towards south direction therefore, Ann`s velocity is 3 mi/h south and her speed is 3 mi/h .

7 0

4 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago

Need a little help here :(

Goshia [24]

Answer:

The output out be 200

Explanation:

Hope this helps :))

8 0

3 years ago

Read 2 more answers