Need help on this thanks

A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ

ludmilkaskok [199]

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

7 0

2 years ago

A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is

Alika [10]

Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

F = G*(M1*M2)/R^2

Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

R = 0.002m

F = 0.0104 N

and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

6 0

2 years ago

3. Rock A is thrown horizontally off of a cliff with a velocity of 20 m/s. The

Anna11 [10]

Answer:

44.1 m

Explanation:

4 0

2 years ago

Due Ma duart ded out 25 N 35 N 1-03

Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

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8 0

2 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

Need help on this thanks​

Need help on this thanks