Ask question

Ask question

All categories

ArbitrLikvidat [17]

3 years ago

6

Franklin needs to ship a box with FedEx. In order to calculate his shipping costs, he needs to measure the mass of the package.

What is the base unit that he should use in this case?
A. meter B. liter C. gram

2 answers:

balandron [24]3 years ago

8 0

Answer:

meter

Explanation:

please mark as brainlist answer

ElenaW [278]3 years ago

4 0

Gram gram gram gram gram

You might be interested in

At an amusement park, a roller coaster carrying nine homosexuals, moving at 18 m/s, jumps its track at a point that is 20 meters

Lorico [155]

It is the same if a white father and a white mother give birth to a black kid, so the mother accuses the father for having (let's say mating) with a black women.

6 0

3 years ago

How does mass affect the gravitational pull of the Sun, Earth, and the Moon?

madreJ [45]

Larger mass creates a stronger pull

3 0

3 years ago

Read 2 more answers

A 1,200 kg dragster, starting from rest, reaches a maximum velocity of 140m/s in 5 seconds. At the 5 second mark, the dragster d

SSSSS [86.1K]

Answer:

Drag or air resistance

Explanation:

The force of friction caused by a moving fluid is called drag. When that fluid is air, it's also known as air resistance.

8 0

3 years ago

People, especially adults, are inactive because of "passive" modes of transportation.

enot [183]

The drop in physical activity is partly due to inaction during leisure time and sedentary behaviour on the job and at home. Likewise, an increase in the use of "passive" modes of transportation also contributes to insufficient physical activity.

3 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0

3 years ago