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3 years ago

How much force is required to accelerate a .6 kg object at 5 m/s^2

Physics

1 answer:

vodomira [7]3 years ago

7 0

Answer:

$.$

» 30 N

Explanation:

Given :

Mass → m → 6kg

Acceleration → a →

$5 {ms}^{ - 2}$

To find :

The formula for force

F = m • a

F = 6 • 5

F = 30 N

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Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t

kompoz [17]

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel $40\times 2=80 miles$

thus its position vector after two hours is

$r_A=-80sin35\hat{i}+80cos35\hat{j}$

similarly B travels with 20 mph and in 2 hours

$=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}$

Thus distance between A and B is

$r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}$

$|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}$

$|r_{AB}|=103.45 miles$

Velocity of A

$v_A=-40sin35\hat{i}+40cos35\hat{j}$

Velocity of B

$v_B=20sin80\hat{i}+20cos80\hat{j}$

Velocity of A w.r.t B

$v_{AB}=v_A-v_B$

$v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}$

4 0

3 years ago

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.

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3 years ago

A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 4.5 cm. The

ANTONII [103]

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = 3.7 rad/s

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

v = (ωs)R = 7.8(65) = 507 cm/s or 5.1 m/s

c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

ωp = 5.38(4.5 cm / 9.5 cm) = 2.55 rad/s

t = θ/ω = 2π / 2.55 = 2.463... 2.5 s

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3 years ago

A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h

Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = $\frac{dr}{dt} X 12 = 4 \frac{ft}{sec} X 12 sec = 48 ft$

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi (48)(4)$

dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

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3 years ago

Why does th ground and atmosphere get warm during the day

Tcecarenko [31]

Answer:

The Greenhouse Effect Revisited. When solar energy strikes the planet during the day, the ground, highways and other objects get hot and absorb that energy. As the sun goes down, the Earth cools by giving off infrared radiation. Because greenhouse gases absorb part of this radiation, the atmosphere warms and keeps the Earth from getting too cold.

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