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3 years ago

The volume of an irregular object can be measured by

Physics

1 answer:

weeeeeb [17]3 years ago

6 0

A.

-because you multiply them all

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A 2 kilogram rabbit starts from rest and is moving at 6m/s after 3 seconds. What net force must be exerted on the rabbit (by the

horrorfan [7]

A = (Vf-Vi) / t,
a = (6-0)/3 = 2m/s^2,

F = ma = 2 * 2 = 4N

8 0

3 years ago

An electric iron is connected to the mains power supply of 220 V. When the electric iron is

Ilya [14]

Answer:

Given: V = 220V, Pmin = 360W, Pmax = 840W

For minimum heating case:

We know that

Pmin = VI

360 = 220 X I

I = 1.63 amp

R = V/I

R = 220/1.63

R = 134.96ohms

For maximum heating case:

We know that

Pmax = VI

840 = 220 X I

I = 3.81 amp

R = V/I

R = 220/3.81

R = 57.74 ohms

7 0

3 years ago

What is the acceleration of a car initially traveling at -5m/s and<br> reaching -22m/s in 3s.

lozanna [386]

Answer:

$a=-5.67\ m/s^2$

Explanation:

Given that,

Initial velocity, u = -5 m/s

Final velocity, v = -22 m/s

Time, t = 3s

We need to find the acceleration of the car. The formula of it is given by :

Acceleration,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(-22)-(-5)}{3}\\\\a=-5.67\ m/s^2$

So, the acceleration of the car is $-5.67\ m/s^2$ .

4 0

3 years ago

How does surface air flow in a middle-latitude cyclone in the northern hemisphere? (1 point)?

Vikki [24]

convergent and counterclockwise

hope it helps :)

3 0

4 years ago

Read 2 more answers

A space vehicle is traveling at 5425 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent

lana66690 [7]

Answer:

$V_{cE}=1489m/s$

Explanation:

Given data

Space vehicle speed=5425 km/h relative to earth

The rocket motor speed=81 km/h and mass 4m

The command has mass m

From the conservation of momentum as the system isolated

$p_{i}=p_{f}\\$

Since the motion in on direction we can drop the unit vector direction

$MV_{i}=4mV_{mE}+mV_{CE}$

Where M is the mass of space vehicle which equals to sum of the motors mass and command mass.

The velocity of the motor relative to the earth equals the velocity of the motor relative to command plus the velocity of the command relative to earth

$V_{mE}=V_{mc}+V_{cE}$

Where Vmc is the velocity of motor relative to command

This yields

$5mV_{i}=4m(V_{mc}+V_{cE})+mV_{cE}\\5V_{i}=4V_{mc}+5V_{cE}$

Substitute the given values

$V_{cE}=\frac{5V_{i}-4V_{mc}}{5}\\ V_{cE}=5425*\frac{1000}{60*60}(m/s)-\frac{4}{5}*81\frac{1000}{60*60}(m/s)\\ V_{cE}=1489m/s$

5 0

4 years ago