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Svetllana [295]
3 years ago
13

A train travels 55 kilometers in 2 hours, and then 52 kilometers in 5 hours. What is its average speed?

Physics
2 answers:
Dafna11 [192]3 years ago
6 0

Average speed = (total distance covered) / (time to cover the distance) .

Total distance covered = (55 km + 52 km) = 107 km

Time to cover the distance = (2 hrs + 5 hrs) = 7 hrs

Average speed = (107 km) / (7 hrs)

Average speed = 15.29 km/hour

bulgar [2K]3 years ago
4 0

The formula for working out speed is distance ÷ time.

55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)

52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)

(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)

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a skinny girl devours food that is equivalent to 600 kcal, as she is anorexic she repents and wants to lose twice as many calori
blondinia [14]

Answer:

Explanation:

Work done in lifting the weight once = mgh

= 20 x 9.8 x (1.9+1.7)

= 705.6 J  

= 705.6 / 4.2 calorie

= 168 cals

Total energy to be spent = 600 x 10³ cals

No of times weight is required to be lifted

= 600 x 10³ / 168

= 3.57  x 10³ times

Total time to be taken = 2 x 3.57 x 10³

= 7.14 x 10³ s

=119 minutes .

4 0
3 years ago
A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red
Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }       ......(1)

Here \lambda_{s} is source wavelength, \lambda_{o} is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, \lambda_{s} = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, \lambda_{0} = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }

\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84

{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056

1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}

\frac{v}{c}=\frac{0.2944}{8.056}

v = 0.036c=0.036\times3\times10^{8}

v = 1.08 x 10⁷ m/s  

8 0
3 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
Select the correct answer.
kipiarov [429]

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

5 0
3 years ago
The path a projectile takes is known as the;
Fantom [35]

B. trajectory

because it is a set projectile wether or not it was calculated

plz mark brainliest

8 0
3 years ago
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