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Evgen [1.6K]
4 years ago
13

What is the velocity of a rocket that goes 700 km north in 25 seconds?

Physics
1 answer:
Svet_ta [14]4 years ago
5 0
Use this formula 
Displacement= velocity x time 
Velocity=Displacement/time 
V=700/25 
V=28 km/s
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In the illustration, which solute will dissolve first?
11Alexandr11 [23.1K]
A cause the sugar cubes are small enough to dissolve fast
6 0
4 years ago
The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?
Gekata [30.6K]

Answer:

The power of the distance is -1.

Explanation:

The equation for the electric potential of a point charge is given by V= k\frac{q}{r}

where V is the electric potential, k is Coulomb's constant (it has a value of 9.0*10^{9} with units \frac{Nm^{2} }{c^{2} }), q is the electric charge of the small charge and r is the distance from the charge.

Now, the power of a number is how many times we multiply that number by itself; we see r appears only once in the equation. So we know the power is 1. But we can see in the equation that k and q are divided by r, which means r is the denominator. This means the power of r is negative (-).

Therefore, the power of r is -1.

5 0
3 years ago
What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of
Murljashka [212]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} . Rearranging the above equation,

A=\frac{\rho l}{R}\\  \frac{\pi d^2}{4}=\frac{\rho l}{R}\\  l=\frac{\pi d^2 R}{4 \rho}

Here d=0.15, R=15, \rho=1.68(10^{-8}).

Substituting numerical values,

l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})}  \\ l=15.78

The length of the wire is 15.78 \;\;m

5 0
4 years ago
Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a
Ira Lisetskai [31]

Answer:

Explanation:

Given

mass of ethanol m_e=310\ gm

mass of aluminium cup m_{al}=90\ gm

both are at an initial temperature of T_i=30^{\circ}C

specific heat of ethanol c_e=2.46\ J/g-K

specific heat of aluminium c_{al}=0.9\ J/g-K

specific heat of ice c_i=2.108\ J/g-K

specific heat of water c_w=4.184\ J/g-K

Latent heat of fusion L=334\ J/gm

suppose m is the mass of ice added

Heat loss by Al cup and ethanol after 18^{\circ}C is reached

Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)

Heat gained by ice such that ice is melted and reached a temperature of 18^{\circ}C

Q_2=m\times 2.108\times (8.5)+m\times 334

Comparing 1 and 2 we get

m=23.65\ gm

Thus 23.65 gm of ice is added

                 

8 0
3 years ago
What is the heart rate recorded a few minutes after completing a workout?
DochEvi [55]
Correct answer should be C
8 0
3 years ago
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