If a net force accelerates a 4.5-kg tool at 40 m/s2, what acceleration would that same net force give to an 18-kg tool?

A block is 10cm long, 5cm wide and 2cm high and weighs 100g. What is the volume of the block? What is the density?

Ivahew [28]

Answer:

1gm/cm^3

Explanation:

its the answer

5 0

3 years ago

What is one advantage of using primary sources when doing research on an

WITCHER [35]

One advantage of using primary sources when doing research on an experiment is, it contain more details about the experiment. Thus, option D is right .

To find the answer, we have to know more about the Primary sources.

<h3>What is the advantages of primary sources?</h3>

Primary sources are the direct or first-hand excerpts that scientists have independently written based on their experiments in order to support their research.
Utilizing original sources encourages analytical and critical thought in relation to the research.

It aids in exploring from various angles, which leads to the discovery of extra facts.
Primary data aids in navigating the conflicts. Since it is a direct resource related to the experiment, it serves as evidence for the data.

Thus, we can conclude that, Primary sources contain more details about the experiment.

Learn more about the Primary sources here:

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5 0

1 year ago

A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy

Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) = fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = $\frac{\alpha }{2} *$ $\frac{\pi r^{2} }{h}$ * a = ( force of gravity ) $\alpha * T\frac{r^{2} }{4} * x * g$

therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

3 years ago

Football player A has a mass of 210 pounds and is running at a rate of 5.0 mi/hr. He collides with player B. Player B has a mass

Naddika [18.5K]

The equation for momentum is p = mv where p is the omentum, m is the mass and v is the velocity. Calculating the momentum for each football player, player A will have a momentum of 1050 lb-mi/h and player B will have a momentum of 570 lb-mi/h. Therefore, momentum of player A is greater than that of player B.

6 0

3 years ago

Read 2 more answers

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)