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An object is moving in a straight line at a constant speed. A resultant force begins to act upon the object. State the ways in w

sergiy2304 [10]

Answer:

accelerate / increase speed OR decelerate / decrease speed OR stop B1

change direction / move in a curve o.w.t.t.e

Explanation:

accelerate / increase speed OR decelerate / decrease speed OR stop B1

change direction / move in a curve o.w.t.t.e

8 0

3 years ago

an object of mass m is rotating about a fixed axis with angular momentum l. its moment of inertia about this axis is i. what is

Tems11 [23]

The Kinetic energy would be 1/2IL².

<h3>What is Rotational Kinetic energy ?</h3>

Rotational energy also known as angular kinetic energy is defined as: The kinetic energy due to the rotation of an object and is part of its total kinetic energy. Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.

As we know linear Kinetic energy = 1/2mv²

where m= mass and v= velocity.

Similarly rotational kinetic energy is given by = 1/2IL²

where I- moment of inertia and L=angular momentum.

To know more about the Kinetic energy , visit:

brainly.com/question/29807121

#SPJ4

8 0

1 year ago

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0

3 years ago

27b. You're in an airplane that flies horizontally with speed 1000 km/h (280 m/s) when an engine falls off. Ignore air resistanc

jonny [76]

Answer:

8400m

Explanation:

The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.

Therefor the horizontal distance the engine travels during its fall is

280 * 30 = 8400m

6 0

3 years ago

135,000 kilometers = how many miles?

DerKrebs [107]

1 mile = 1.609 km

(135,000 km) x (1 mile / 1.609 km) = 83,885.1 miles

5 0

3 years ago