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Pepsi [2]
1 year ago
10

A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d

irection at a velocity of +4 meters/second with respect to himself. If the ball lands after 30 seconds, about how far does it land horizontally from the point at which it was released?
A. 120 meters B. 125 meters C. 300 meters D. 420 meters
Physics
1 answer:
yulyashka [42]1 year ago
3 0

Option (D) is correct.

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Explanation:

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

d= 420 m

Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

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Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume
REY [17]

Explanation:

The given data is as follows.

          mass, m = 75 g

      T_{1} = 0^{o}C

      T_{2} = 27^{o}C

      Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

            q = m C \times (T_{1} - T_{2})

               = 75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C

               = 8464.5 J/mol

               = 8.46 kJ ......... (1)

Also, it is given that T_{3} = -20^{o}C = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

        q = mC \times (T_{3} - T_{1})

           = 0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K

           = -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

               8.46 kJ - 46.32 kJ

             = -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0^{o}C to ice at -20.0^{o}C is -37.86 kJ.

4 0
2 years ago
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
KIM [24]

Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

       t_1 =  250 \ s

The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

7 0
2 years ago
A race car driver must average 200km/hr for four laps to qualify for a race. Because of engine trouble, the car averages only 17
vampirchik [111]
The average speed would have to be 260 km/hr due to the driver originally going 30 km/hr too slow the first two laps
5 0
2 years ago
A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the
UkoKoshka [18]

Answer:

 Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, KE = \frac{1}{2} mv^2, where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

  1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg

So, mass of bike = 38 kg.

3 0
2 years ago
A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds
solong [7]

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/s^{2}

B. Formula: a=\frac{V-V_{0} }{t} and s=d/t

speed= 394.6/15

s=26.3m/s

a=\frac{26.3-0}{15}

a=1.75m/s^{2}

C. 7.9-1.75=difference of 6.15m/s^{2}

D. The force that most likely caused this difference is friction forces

3 0
1 year ago
Read 2 more answers
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