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4 years ago

How many neutrons does element x have if it’s atomic number is 23 and it’s mass is 90

Physics

1 answer:

Oxana [17]4 years ago

3 0

The amount of neutrons would be 23.

The atomic number represents the amount of neutrons, so all you have to do is look at it to know. The atomic mass was put here simply to mislead you. B.T.W, the element is Valadium.

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A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo

guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s $(1km\h=\frac{5}{18} m/s)$

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

$33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)$

$a = 1.83 m/s^2$

$Force=mass\times acceleration$

$Force=710 kg\times 1.83 m/s^2$

$Force=1,299.3 N$

The net force acting on the car 1,299.3 N.

8 0

3 years ago

juie ran eight laps around her school's quarter mile track in 18 minutes. her average velocity was _______m/s

Blizzard [7]

Answer:

About 3m/s

Explanation:

If Julie ran eight quarter mile laps, then she ran a total of 1/4 * 8=2 miles. 2 miles is about 3220 meters, and 18 minutes is 1080 seconds, meaning that her average velocity was about 3m/s. Hope this helps!

5 0

3 years ago

Read 2 more answers

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

3 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

3 years ago

High speed wind current is called

rusak2 [61]

High speed wind current is called ocean current

6 0

3 years ago

Read 2 more answers