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2 years ago

You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

Physics

1 answer:

bagirrra123 [75]2 years ago

3 0

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60 hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

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If a drop is to be deflected a distance d = 0.350 mm by the time it reaches the end of the deflection plate, what magnitude of c

Sloan [31]

Answer:

q = 4.87 X 10^ -14 C

Explanation:

As d=0.350 mm

The ink drop will be accelerated by the electric field between the plates:

a = F/m

d = a(D0 / v)^2 / 2 ...... 1

a = qE/m ............... 2

Substituting 2 into 1:

d = (qE/m)(D0 / v)^2 / 2

q = 2mdv^2 / [E(D0)^2]

q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]

q = 4.87e-14 C

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2 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

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2 years ago

Which of the following is the BEST example of increasing the intensity of a workout? A. running one mile further than normal B.

stepladder [879]

Answer:

B. running one mile faster than normal

Explanation:

4 0

2 years ago

Read 2 more answers

You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba

Cloud [144]

Answer:

The ball reaches Barney head in $t = 8 \ s$

Explanation:

From the question we are told that

The rise velocity is $v = 14.70 \ m/s$

The height considered is $h = 196 \ m$

The horizontal velocity of the large object is $v_h = 8.50 \ m/s$

Generally from kinematic equation

$s = ut + \frac{1}{2} gt^2$

Here s is the distance of the object from Barney head ,

u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

$u = -14.7 m/s$

$196 = -14.7 t + \frac{1}{2} * 9.8 * t^2$

= $4.9 t^2 - 14.7t - 196 = 0$

Solving the above equation using quadratic formula

The value of t obtained is $t = 8 \ s$

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2 years ago

The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?

Fynjy0 [20]

The formula for potential energy is PE=mgh
It can have that high of a potential energy if it's relative height what super high.

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2 years ago