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3 years ago

A car travels 10 miles East in 30 minutes. What is its velocity in miles per hour?

Physics

2 answers:

Stella [2.4K]3 years ago

8 0

Answer:

Velocity = 20 mph

Explanation:

Distance = 10 miles

Time = 30 min = 0.5 hour

Velocity = distance/time

= 10miles/0.5 hour

= 20 mph

Ghella [55]3 years ago

7 0

The car's velocity is

20 miles per hour East.

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234 oz to tons using the table method (US)

Rom4ik [11]

Answer:

234 oz = 0.0066339 t.

Explanation:

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3 years ago

Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri

Artemon [7]

First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

Ep=m*g*h=0.7*9.8*7=48.02 J

The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick

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3 years ago

Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo

luda_lava [24]

Answer:

Explanation:

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2 years ago

What is a non contact force that attracts all objects to the centre of the earth

beks73 [17]

Answer:

Gravity. The weight-force or weight of an object is the force because of Gravity, which acts on the object attracting it towards the centre of the earth.

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3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

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2 years ago