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2 years ago

Given a force of 10 N and an acceleration of 5 m/s 2. What is the mass ?

Mathematics

2 answers:

Airida [17]2 years ago

7 0

Solution,

Here,

Given,

Force= 10N

Acceleration= 5m/s^2

Mass= ?

Therefore, as we know that

Force= Mass× Acceleration

10=M×5

10/5= M

Therefore, M= 5kg.

<h2>If it helped, please mark it as the brainliest answer :)</h2>

Ymorist [56]2 years ago

3 0

Answer:

2 kg.

Step-by-step explanation:

F = m * a (Newton's Second Law of Motion).

10 = m * 5

The mass m = 10/5 = 2 kg.

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Destiny paid $6.60 for a dozen cupcakes for a birthday patry. What is the unit cost for each cupcake?

Digiron [165]

Answer: $.55

Step-by-step explanation:

An important thing to know is that a dozen is equal to 12.

12x= 6.60

/12 /12

x=.55

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3 years ago

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300-7 [4 (3+5)]+3 to the 3rd power

Allisa [31]

The expression to solve is the
300-7 [4 (3+5)] + 3 to the 3rd power
3 to the third power means 3³, so
300-7 [4(3+5)]+3³
= 300 - 7 [4(8)] + 27
= 300 - 7[32] + 27
= 300 - 224 + 27
= 76 + 27
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so, by solving this we get 103

4 0

3 years ago

What is the volume??????

Ugo [173]

Answer:

$36\:\mathrm{in^3}$

Step-by-step explanation:

The area of a prism is given by the product of the base and the height. In this triangular prism, the area of the either base is:

$A=\frac{1}{2}\cdot 3\cdot 4=6\:\mathrm{in^2}$

The height is marked as $6$ inches. Thus, the volume of the prism is:

$\frac{1}{2}\cdot 3\cdot 4\cdot 6=\boxed{36\:\mathrm{in^3}}$

7 0

3 years ago

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BRAINLIEST WILL BE AWARDED PLEASE HELP

o-na [289]

X=3
Just look at where the two lines intercept and see the x value

5 0

3 years ago

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Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago