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Rus_ich [418]
3 years ago
11

6th grade science !!!!!!!!!!!!!!!​

Physics
1 answer:
posledela3 years ago
5 0

Answer:

71

Explanation:

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sammy [17]

Answer:

mesuring heigh and weight is important

3 0
3 years ago
What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic field
goldenfox [79]
F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
6 0
3 years ago
Help! (Look in the pic)
hjlf

Answer:

turtle

Explanation:

5 0
3 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
3 years ago
A 0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of co
Alex_Xolod [135]

To solve this problem, we should recall the law of conservation of energy. That is, the heat lost by the aluminium must be equal to the heat gained by the cold water. This is expressed in change in enthalpies therefore:

- ΔH aluminium = ΔH water

where ΔH = m Cp (T2 – T1)

The negative sign simply means heat is lost. Therefore we calculate for the mass of water (m):

- 0.5 (900) (20 – 200) = m (4186) (20 – 0)

m = 0.9675 kg

 

Using same mass of water and initial temperature, the final temperature T of a 1.0 kg aluminium block is:

- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)

- 900 T + 180,000 = 4050 T

4950 T = 180,000

T = 36.36°C

 

The final temperature of the water and block is 36.36°C

4 0
4 years ago
Read 2 more answers
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