3 years ago

One way to decrease the inertia of an object is to

Physics

2 answers:

ANTONII [103]3 years ago

8 0

The only way to decrease the inertia of an object is to remove some of its mass. (B)

kipiarov [429]3 years ago

7 0

B.

I need to fulfill a character requirement so...$1&:&/&2&&:&;& kwkdndksjwken fjwown fnfj

You might be interested in

Which type of mirror causes light to spread out?

saw5 [17]

A convex mirror makes a reflected light rays spread out.

3 0

3 years ago

Read 2 more answers

Which of the following is a force acting between objects that do not touch?

kirill [66]

The answer is B friction force

3 0

3 years ago

Read 2 more answers

One of the best ways to avoid injury while riding a bike

mel-nik [20]

know how to ride a bike simply put also wear safety gear like a helmet knee pads things like that

hope this helps

5 0

2 years ago

Read 2 more answers

Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by

bixtya [17]

Answer:

v = 0.99 c = 2.99 x 10⁸ m/s

Explanation:

From the special theory of relativity:

$t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2} } }\\$

where,

v = speed of travel = ?

c = speed of light = 3 x 10⁸ m/s

t = time measured on earth = 90 years

t₀ = time measured in moving frame = 6 months = 0.5 year

Therefore,

$90\ yr = \frac{0.5\ yr}{\sqrt{1-\frac{v^2}{c^2} } } \\\\\\\sqrt{1-\frac{v^2}{c^2}} = \frac{0.5\ yr}{90\ yr}\\ 1-\frac{v^2}{c^2} = 0.00003086\\\frac{v^2}{c^2} = 1-0.00003086\\$

8 0

3 years ago

A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05

7nadin3 [17]

Answer:

$W_{drag} = 4.223\,J$

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

$U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}$

The work done on the ball due to drag is:

$W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})$

$W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})$

$W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]$

$W_{drag} = 4.223\,J$

7 0

3 years ago

Read 2 more answers