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mafiozo [28]
3 years ago
5

when an object is moving there is always a force pulling it in the direction of its motion?true or false? explain

Physics
1 answer:
pickupchik [31]3 years ago
8 0
This is false because an object can also be pushed by force and isn't always have to be pulled by a force.
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The charge on the moving particle
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1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00
rosijanka [135]

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

8 0
2 years ago
A periodic wave with wavelength 2m has a speed of 4m/s. What is the waves frequency?.
Gekata [30.6K]

The wave frequency is 2 Hz.

What is wave frequency ?

The number of waves that pass through a fixed point in a given amount of time is referred to as the wave frequency. The hertz is the SI unit for wave frequency (Hz).

f = v / w

where,

f = frequency\\v = speed \\w = wavelength

Given,

v = 4 m/s, w = 2 m

f = 4/2 \\f = 2 Hz\\

The waves frequency is 2 Hz.

To know more about wave frequency,check out:

brainly.com/question/15830195

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5 0
1 year ago
Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 1
Novosadov [1.4K]
<span>they are travelling at right angles to each other.
 At any given instant they form a right triangle with their starting point
 </span>South bound <span>= x  [mi/h]
</span> East bound <span> = x+1 [mi/h]
 after five hours they will be
 d=5x
 and
 d=5(x+1)
 miles away from the starting point 
 (5x)^2+(5(x+1))^2=625
 25x^2+(5x+5)^2=625
 25x^2+25x^2+50x+25=625
 50</span>x^2+50x-600=0
<span> x^2+ x - 12=0
 (x+4)(x-3)=0
 take the postive value
 x= 3 mph the speed of south bound
 4mph east bound </span>
6 0
3 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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