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3 years ago

1. How much force (N) is required to accelerate a 2.60 kg remote controlled toy car at a rate of 8 m/s2?

Mathematics

1 answer:

Natali [406]3 years ago

8 0

Answer:

Step-by-step explanation:

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I'm supposed to write out an algebraic equation, and then solve the equation. If you can help me, with work shown, you'll get vo

Arada [10]

Answer:

simple just follow me and mark it brainliest

Step-by-step explanation:

LET THE THREE CONSECUTIVE EVEN INTEGERS BE

(x+2) ,(x+4) ,(x+6)

then

according to the question

4(x+2) – (x+6) = 2(x+4) + 6

4x + 8 – x – 6= 2x + 8+6

3x +2 = 2x + 14

3x – 2x= 14 –2

x= 12

the numbers are

5 0

3 years ago

Read 2 more answers

In a sample of 77 children , the mean age as at which they first began to combine words was 16.51 months

astraxan [27]

Answer:

Mean age = 1271.27

Step-by-step explanation:

No of children = 77

The age at which children first began to combine words = 16.51 months

Mean age = No of children × age at which children first began to combine words

= 77 × 16.51

= 1271.27

Mean age = 1271.27

7 0

3 years ago

This graph represents

hram777 [196]

It is C (2, 1) I think it is because they intersect

8 0

2 years ago

-2x

mars1129 [50]

y=0

................your welcome

6 0

3 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago