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Phantasy [73]
2 years ago
13

How does the ratio of tin to copper affect the properties of the alloy bronze?

Physics
1 answer:
Pani-rosa [81]2 years ago
7 0

Answer:

Stronger and harder than either of the pure metals

Explanation:

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Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
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.
dsp73

Answer:

anyone know this or should i get my brother

7 0
2 years ago
Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
Marta_Voda [28]

Answer:

Acceleration, a=9.91\times 10^{15}\ m/s^2

Explanation:

It is given that,

Separation between the protons, r=3.73\ nm=3.73\times 10^{-9}\ m

Charge on protons, q=1.6\times 10^{-19}\ C

Mass of protons, m=1.67\times 10^{-27}\ kg

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

ma=\dfrac{kq^2}{r^2}

a=\dfrac{kq^2}{mr^2}      

a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}      

a=9.91\times 10^{15}\ m/s^2

So, the acceleration of two isolated protons is 9.91\times 10^{15}\ m/s^2. Hence, this is the required solution.

3 0
3 years ago
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
3 years ago
Do acids or bases turn litmus paper red?
Tom [10]
Base => Blue

aciD => reD
5 0
3 years ago
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