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Murljashka [212]

3 years ago

9

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN

KS!!! What is the fundamental frequency of a mandolin string that is 42.0 cm long when the speed of waves of the string is 329 m/s? with working outs

1 answer:

Juliette [100K]3 years ago

7 0

Answer:

f = V/λ = 329/0.42 = 783 Hz

:)

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3 years ago

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After the last major ice age, sea levels around the world rose. As they did, the rising waters divided what was once a single is

beks73 [17]

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3 years ago

A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr

GrogVix [38]

Answer:

The speed of the block when is fallen 30cm

v=0.726 $\frac{m}{s}$

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ = μN = μ(m1)g

$(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a$

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

$(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a$

$0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }$

$v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}$

5 0

4 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim

miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
We can find this distance simply applying the definition of average velocity, as follows:

$\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)$

The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

$v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)$

where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
Solving for Δx, we get:

$\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)$

So, the total distance traveled was the sum of (1) and (3):
Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

$\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)$

Δx₂, is the distance traveled while decelerating, and can be obtained using (2):

$v_{omax} ^{2} = 2* a* \Delta x_{2} (7)$

Solving for Δx₂, we get:

$\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)$

Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
Taking the positive root in the quadratic formula, we get the following value for vomax:
v₀max = 24.3 m/s.

6 0

3 years ago

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radi

guapka [62]

Answer:

(A) E = 2.633 * 10^19 N/C

(B) E = 1.096 * 10^13 N/C

(C) E = -1.096 * 10^13 N/C

Explanation:

Parameters given:

Number of protons, N = 92

Radius of Uranium nucleus = 7.4 * 10^-15 m

Electronic charge, ẹ = 1.6023 * 10^-19 C

Electric field at a point R due to a charge Q is given as

E = (k*Q) / (R^2)

Where k =Coulombs constant

(A) Since there are 92protons,the total Electric field due to the protons will be:

E = (k*e*N) / (R * R)

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (7.4 * 10^-15)^2

E = 2.633 * 10^19 N/C

(B) At the position of the electrons, R = 1.1 * 10^-10m. Therefore, Electric field will be:

E = (9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = 1.096 * 10^13 N/C

(C) There are 92 electrons in the Uranium atom and electrons have a charge - e, hence, the Electric field due to the electrons at the nucleus will be:

E = -(k*e*N) / (R * R)

E = -(9 * 10^9 * 92 * 1.6023 * 10^-19) / (1.1 * 10^-10)^2

E = -1.096 * 10^13 N/C

7 0

3 years ago