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sammy [17]
3 years ago
13

The greater the mass of an object? I’ll give brainliest!*

Physics
1 answer:
liraira [26]3 years ago
6 0

between B and C im thinking its more C though

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What is the term for the overall change in position?
mart [117]
That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.
6 0
3 years ago
Read 2 more answers
A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2
lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

7 0
3 years ago
A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.
Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

\omega =\frac{2\pi \times 469}{60}=49.11 rad/s

t=16 s

Angular deceleration in 16 s

\omega =\omega _0+\alpha \cdot t

\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2

Moment of Inertia I=mr^2=13\times 1.8^2=42.12 kg-m^2

Change in kinetic energy =Work done

Change in kinetic Energy=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}

\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J

(a)Work done =50.79 kJ

(b)Average Power

P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW

7 0
3 years ago
What mass of silver (in grams) is solidified when 749 joules of heat are released by a sample of molten silver at its freezing p
Anastasy [175]

heat released Q = 749 joules

heat of fusion of silver L = 109 J/g

Here phase of silver is changing from liquid to solid

so temperature will remain same

all heat will be released due to its phase change

and in this case we use Q=mL

where m is the mass of silver in gram

Q= mL

749 = m * 109

m = 749/109

m = 6.87 gram

4 0
3 years ago
Read 2 more answers
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