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1 month ago

What is the mass of a baseball clocked moving at a speed of 105 mph or 46.9 m/s and wavelength 9.74 × 10-35m?

1 answer:

6 0

To address this issue, we apply the de Broglie equation written as:

λ = h/mv
where h equals 6.626×10⁻³⁴ J·s

Solving for m, we substitute for v, which is 46.9 m/s:
9.74 × 10⁻³⁵ m = 6.626×10⁻³⁴ J·s / (m)(46.9 m/s)
Thus, we find that m = 0.145kg

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A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

Yuliya22 [3333]

Answer:

The box experiences an acceleration with a magnitude of 2 m/s².

Explanation:

Details provided:

Mass of the box: $m=5.0$ kg

Force acting towards the east: $F=27$ N

Frictional force acting towards the west: $f=17$ N

Let’s denote the acceleration as $a$ m/s².

Thus, the net force acting on the box in the east direction is stated as:

$F_{net}=F-f=27-17=10\textrm{ N}$

According to Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Thus, the magnitude of the box's acceleration is determined to be 2 m/s².

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1 month ago

Read 2 more answers

Sebuah benda dijatuhkan bebas dari ketinggian 200 m jika grafitasi setempat 10 m/s maka hitunglah kecepatan dan ketinggian benda

inna [3103]

Kindly use English in your post so that I or someone else can assist you.

7 0

1 month ago

Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.

serg [3582]

Answer:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

Consequently, we find that:

$V_A = \frac{1}{4} V_B$

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

$L_A = L_B =L$ both wires share the same length

both wires carry an identical current $I_A = I_B =I$

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

$n_A = n_B =n$

We also know that $r_A = 2 r_B$ where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

Thus, we conclude that

$A_A = 4 A_B$

Now we are aware that the drift velocity of an electron in a wire can be described by:

$v_d = \frac{I}{neA}$

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

So we observe that:

$V_A = \frac{1}{4} V_B$

Thus, the most fitting answer is:

a. vA = vB/4

6 0

25 days ago

3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta

ValentinkaMS [3465]

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

6 0

1 month ago

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

Yuliya22 [3333]

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (we take downward direction as positive)

$v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s$ Using the equations of motion

The pebble's speed upon contact with the water is 19.62 ms

3 0

16 days ago