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kenny6666 [7]
3 years ago
13

An oscillating block-spring system has a mechanical energy of 4.72 J, an amplitude of 11.2 cm, and a maximum speed of 4.24 m/s.

Find (a) the spring constant, (b) the mass of the block and (c) the frequency of oscillation.
Physics
1 answer:
dsp733 years ago
7 0

Answer:

anyone know this or should i get my brother

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A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50
Lelechka [254]

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

It takes 0.21 sec to reach the bottom from apartment

So

s=u_1t+\frac{gt^2}{2}

1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}

u_1=6.11 m/s

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

So height of upper floor above window

v^2-u^2=2as

where s= height of upper floor above window

here u=0

6.11^2=2\times 9.81\times s

s=\frac{6.11^2}{2\times 9.81}

s=1.902 m

5 0
3 years ago
A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times
Ira Lisetskai [31]

Answer:

9R

Explanation:

We know that the resistance is R=\rho *\frac{L}{A}.

If we stretch the wire to a new length L2 = 3L, the cross-sectional area will also change. If the cross-sectional area doesn't change throughout the wire, we can say that:

Volume = L*A = 3L * A2    being A2 the new area after stretching the wire.

Since the volume remains the same we conclude that A2 = A/3

With this information, we calculate the new resistance:

R2=\rho *\frac{L2}{A2}=\rho *\frac{3*L}{A/3}=\rho * 9 * \frac{L}{A}

Since R=\rho *\frac{L}{A}, and by simple inspection of the previous equation, we get:

<em>R2 = 9*R</em>

6 0
3 years ago
Select the correct answer
Sergeu [11.5K]

Answer:

I may be a little late, but it's A i got it right on the exam

Explanation:

5 0
2 years ago
A thin, metallic spherical shell of radius 0.187 m has a total charge of 6.53×10−6 C placed on it. A point charge of 5.15×10−6 C
MAVERICK [17]

Answer: a) 112.88 * 10^3 N/C; b) The electric field point outward from the center of the sphere.

Explanation: In order to solve this problem we have to use the gaussian law so we use a gaussian surface at r=0.965 m and the electric flux is equal to Q inside/εo

E* 4*π*r^2= Q inside/εo

E= k*Q inside/r^2= 9*10^9*(6.53+5.15)μC/(0.965)^2=122.88 * 10 ^3 N/C

5 0
3 years ago
A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com
blagie [28]

Answer:

4500.5 nutritional calories per gram

Explanation:

Heat lost by the new candy = heat gained by the bomb calorimeter.

Heat gained by the bomb calorimeter = c×ΔT

where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K

ΔT = change in temperature = 2.69°C = 2.69 K.

Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J

Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

8 0
3 years ago
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