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3 years ago

How many ft/s is 60 mph/hr

Physics

1 answer:

olga nikolaevna [1]3 years ago

3 0

Answer:

1 Mile per hour = 1.60934 kilometers per hour = 0.44704 meters per second (SI base unit). 1 mph = 0.447 04 m/s. 1 Foot per Second: Feet per Second.

Explanation:

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A tugboat pulls a ship with a constant net horizontal force of 4.71 × 103 N and causes the ship to move through a harbor. How mu

Studentka2010 [4]

Answer:

21.53 x 10^{6} N

Explanation:

force (F) = 4.71 x 10^{3} N

distance (s) = 4.57 km = 4570 m

how much work is done.

work = force x distance

work = 4.71 x 10^{3} x 4570 = 21.53 x 10^{6} N

8 0

3 years ago

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy.
Potential energy.

Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$

$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0

3 years ago

Hey stob it.<br> Please help me.<br> Cmon help me.<br> Plz.

Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

6 0

2 years ago

Type the correct answer in the box. Round your answer to the nearest whole number. Calculate the man’s mass. (Use PE = m × g × h

Blababa [14]

Answer:

56 kg

Explanation:

The change in potential energy of the man is given by:

$\Delta U = mg \Delta h$

where

m is the man's mass

g is the gravitational acceleration

$\Delta h$ is the change in height of the man

In this problem, we have:

$\Delta U=4620 J$ is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

$\Delta h=8.4 m$ is the change in height

Re-arranging the equation and substituting the numbers, we find the mass:

$m=\frac{\Delta U}{g\Delta h}=\frac{4620 J}{(9.8 m/s^2)(8.4 m)}=56 kg$

6 0

3 years ago

Read 2 more answers

A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. A) cm/h B) cm/min C) m/h

Crazy boy [7]

A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h

7 0

3 years ago