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2 years ago

How much force is needed to accelerate a 4kg mass at 12m/s/s?

Physics

2 answers:

zloy xaker [14]2 years ago

7 0

I guess 48 , but I’m not sure

lions [1.4K]2 years ago

4 0

Force=mass x acceleration
=4 x 12
= 48N

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A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi

Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of $12-2x$

Dimension of new box is $(12-2x)\times (12-2x)\times x$

Volume $V=(12-2x)\times (12-2x)\times x$

$V=\left ( 12-2x\right )^2\cdot x$

For maximum volume differentiate with respect to x we get

$\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0$

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

$V=\left ( 12-2\cdot 4\right )^2\cdot 4$

$V=4^3$

$V=64\ in.^3$

6 0

3 years ago

Which restricted basin has the coolest temperatures?

sattari [20]

Hudson Bay is the restricted basin that has the coolest temperatures

Hudson Bay is a restricted basin which remains frozen or is dominated by ice over the summer solstice and through- out much of the high-sun season. This basin experiences a harsh continental climate.

The average annual temperature in almost the entire bay is around 0 °C (32 °F) or below. In the extreme northeast, winter temperatures average as low as −29 °C or −20.2 °F. The region of this basin has very low year-round average temperatures.

This basin starts freezing up by early November, and the northern part of the basin is typically entirely iced over by the end of the month.

correct answer is Hudson bay

learn more about basin :

brainly.com/question/11871406?referrer=searchResults

#SPJ4

3 0

2 years ago

Analyze the chemical reaction below

34kurt

Answer:

Please find the answer in the explanation.

Explanation:

Given that 16 g CH4 + 64 g 02 - 44 g CO2 + 36g H2O

To explain the law of conservation of mass and describe how the equation represents the law of conservation of mass, let me first start from law.

The law state that: mass can neither be created nor destroyed.

The mass of each element at the reaction side must be equal or the Same with the magnitude of mass at the product

The equation represents the law of conservation of mass because the mass of molecules at the right hand side is equal to or balance with the molecules at the left hand side. For example, the number of Oxygen, and othe elements are the at both side.

6 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago

The air in a tire pump has a volume of 1.5 L at a temperature of 5 ℃. If the temperature is increased to 25 ℃ and the pressure r

tankabanditka [31]

Answer:

Explanation:

5 C = 278 K

25 C = 298 K

V1 / T1 = V2 / T2

1.5L / 278 K = V2 / 298 K

V2 = (1.5L * 298) / 278

V2 = 1.61 L

5 0

3 years ago