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kogti [31]
2 years ago
9

a planes average speed between two cities is 600 km/hr. if he takes 2.5 hrs. how far does the plane fly

Physics
1 answer:
o-na [289]2 years ago
4 0

Answer:

1500 hr

..............

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Wind is the result of ___ currents.
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The word is Conventional.
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A proton travels with a speed of 5.05 ✕ 106 m/s at an angle of 64° with the direction of a magnetic field of magnitude 0.160 T i
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Answer:

A. 1.19 * 10^(-13) N

B. 7.12 * 10^(15) m/s²

Explanation:

Parameters given:

Speed, v = 5.05 * 10^6 m/s

Angle, A = 64°

Magnetic field strength, B = 0.160T

Mass of proton, m = 1.673 * 10^(-27) kg

Charge of proton, q = 1.6023 * 10^(-19) C

A. Magnetic force is given as:

F = q*v*B*sinA

F = 1.6023 * 10^(-19) * 5.05 * 10^6 * 0.160 * sin64

F = 1.19 * 10^(-13) N

B. Force is generally given as:

F = m*a

Hence, we can find acceleration, a, by making it the subject of formula:

a = F/m

a = (1.19 * 10^(-13))/(1.673 * 10^-27)

a = 7.12 * 10^15 m/s²

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3 years ago
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All objects emit what radiation
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If an object is not at Absolute Zero, then it is
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A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s
valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

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acceleration, a, is provided for x seconds to increase the speed from 1 m/s to  3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2.  The cart increases speed by 2.5 m/s every second.  

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

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2 years ago
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
3 years ago
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