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zmey [24]
3 years ago
9

Which of these is a characteristic of climate

Physics
1 answer:
deff fn [24]3 years ago
7 0

Answer:

No picture

Explanation:

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7. A bicycle accelerates at 2.0 m/s2. If the mass of the bicycle and rider together is
kkurt [141]

Answer:

170 N

Explanation:

Since Force F = ma were m = mass = 85 kg and a = acceleration = 2.0 m/s².

So the net force on the bicycle is

F = ma = 85 kg × 2.0 m/s² = 170 N

4 0
3 years ago
Your average speed on the first half of a car trip is 69.0 km/h. How fast do you have to drive on the second half of the trip to
vova2212 [387]

Answer:

13 km/h

Explanation:

Average speed = distance/time

Let the total distance and total time taken for the whole trip be d km and t hours respectively

Average speed for the whole trip = 82 km/h

d = 82t

The distance covered in the first half = d1/2

Time taken = t/2

Average speed = 69 km/h

69 = d1/2 ÷ t/2

d1 = 69t

The distance covered in the second half = d2/2

Time taken = t/2

Let the average sly for the see half be A

A = d2/2 ÷ t/2

d2 = At

d = d1 + d2

82t = 69t + At

At = 82t - 69t

At = 13t

A = 13t/t = 13 km/h

3 0
4 years ago
The boy on the tower throws a ball 20m downrange. What is his pitching speed?
san4es73 [151]

Answer:

The pitching speed of the ball is 19.7 m/s

Explanation:

  • Here, we can use the third equation of motion,  v^{2} = u^{2} - 2as
  • whereas v represents the final velocity, u represents initial velocity, a is the acceleration due to gravity and s is the displacement or distance an object traveled
  • Here, the initial velocity of the the ball is given as  zero and the acceleration due to gravity is 9.8  , the distance 's' is given as 20 m
  • Using the equation,  v^{2} = 2 * 9.8 * 20 = 392\\v = \sqrt{392} = 19.7m/s
  • Hence, the pitching speed of the ball is 19.7 m/s

5 0
3 years ago
I need help with this question
Juli2301 [7.4K]

Answer:

Centripetal force is the force that keeps the yoyo going in a circle, if the string breaks, the yoyo would would fly off in a direction that is different to the point on the circle.

8 0
3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
3 years ago
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