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garri49 [273]
2 years ago
7

Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a

nd one on the Moon. Which of the following three statements is/are true? I. The horizontal distance traveled by the bullet is greater for the Moon
Physics
1 answer:
olga nikolaevna [1]2 years ago
3 0

Answer:

horizontal distance on moon > horizontal distance on earth

Explanation:

Let the time of fall is t and the height from which they fall is h.

The time of fall is

t=\sqrt{\frac{2h}{g}}

Time of fall on earth

t=\sqrt{\frac{2h}{g_{e}}

Time of fall on moon

t'=\sqrt{\frac{2h}{g_{m}}

As the gravity on moon is less than the gravity on earth

So, t' > t

Horizontal distance = horizontal velocity x time

So, the horizontal distance on moon is more than the horizontal distance on earth.

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A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is t
Korolek [52]

Answer:

static friction acting opposite to the direction of travel

Explanation:

Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.

Explanation:

8 0
2 years ago
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
kvasek [131]

<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

7 0
2 years ago
A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal
ASHA 777 [7]
Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
F_{centripetal\:force} =  \frac{m*s^2}{R}

Solving:
F_{centripetal\:force} = \frac{m*s^2}{R}
F_{centripetal\:force} = \frac{68*3.9^2}{6.5}
F_{centripetal\:force} = \frac{68*15.21}{6.5}
F_{centripetal\:force} = \frac{1034.28}{6.5}
\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}
Answer:
<span>B.159 N</span>
3 0
2 years ago
What phenomenon does this image demonstrate? *
Ostrovityanka [42]

Answer:

Constructive Interference

Explanation:

8 0
2 years ago
Read 2 more answers
A student states that 203.9 g of table sugar will form a saturated solution in 100 g of water.
riadik2000 [5.3K]

The temperature and the solubility of sugar at that temperature

Explanation:

The amount of substance which can be dissolved in the solvent depends on the temperature.

As the temperature increases, more substance can be dissolved.

A solution is  saturated if any more of the solute cannot be dissolved in the solution at the given temperature

Hence we need to know the temperature  and also the amount of substance which can be dissolved(solubility) at the same temperature

a) the statement given in option A is correct

b) molar mass has no correlation with the substance's solubility and hence option b is not correct

c) The percent by volume of the solution is not needed to find if the solution is saturated and hence option c is not correct

3 0
2 years ago
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