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2 years ago

What is the momentum of a 1, 500 kg car traveling at 3m/s?

Physics

2 answers:

ruslelena [56]2 years ago

6 0

4,500 kgm/s bc 1,500x3

marusya05 [52]2 years ago

3 0

<h3>Answer: D) 4,500 kg m/s</h3>

========================================

Work Shown:

To find the momentum, we multiply the mass and the velocity.

m = mass = 1500 kg

v = velocity = 3 m/s

p = momentum

p = m*v

p = (1500 kg)*(3 m/s)

p = (1500*3) kg m/s

p = 4500 kg m/s

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What is the resultant of a pair of one pound forces at right angles to each other?

OleMash [197]

Answer:

$F_R=\sqrt{2} \ pound.force$

Explanation:

Given that there are two force of 1 pound each at right angles to each other.

The from the vector law of addition:

$F_R=\sqrt{F_1^2+F_2^2}$

where:

$F_R=$ resultant force

$F_1\ \&\ F_2$ be the two of the forces to be added.

$F_R=\sqrt{1^2+1^2}$

$F_R=\sqrt{2} \ pound.force$

8 0

3 years ago

An alternating current of 60hz changes direction 100 times per second.

Dvinal [7]

Answer:

False

Explanation:

Because it is 60 hz

7 0

1 year ago

A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement

Igoryamba

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

$\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}$

The angular displacement is given by:

$\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}$

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

$\displaystyle \alpha=\frac{6-13.5}{7}$

$\displaystyle \alpha=\frac{-7.5}{7}$

$\displaystyle \alpha=-1.071 \ rad/s^2$

Thus, the angular displacement is:

$\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}$

$\displaystyle \theta=94.5-26.25$

$\boxed{\displaystyle \theta=68.25\ rad}$

Angular displacement=68.25 rad

3 0

3 years ago

7 km is equal to:<br><br> 0.7 m<br> 70 m<br> 700 m<br> 7,000 m

dusya [7]

The answer is 7000 meters.

7 0

3 years ago

Read 2 more answers

A tightrope walker is walking between two buildings holding a pole with length L=14.0 m, and mass mp=17.5 kg. The daredevil grip

Artemon [7]

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

F_left = F_right = F

W_bird L / 2 - 2 F 0.595 = 0

F = $\frac{m_{bird} \ g L} { 4 \ 0.595}$

we calculate

F = 0.560 9.8 14.0 /2.38

F = 32.28 N

7 0

3 years ago