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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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How to Solve proposition?

Helen [10]

Do you mean proportions?

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3 years ago

Which two points lie on the circumference of this circle? A. (1.3,0.5) B. (2.5,1.5) C. (3.4, -0.2) D. (3.9,0.7) E. (6.0,-2.0)

liq [111]

Step-by-step explanation:

(x-2)²+(y+5)²=25

option C (3.4,-0.2)

LHS= (3.4-2)²+(-0.2+5)²

= (1.4)²+(4.8)²

= 1.96+23.04

= 25

option E(6,-2)

LHS= (6-2)²+(-2+5)² = 4²+(-3)² = 16+9 = 25

6 0

3 years ago

If m 6=51, then m 7=

Sophie [7]

Answer:

m 7=129

Step-by-step explanation:

if 6 and 7 are right next to each other. it makes 180 degrees so you would take 180-51 to get your answer

6 0

3 years ago

Pl help it’s for homework

Reika [66]

It’s D for your answer

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3 years ago

The sum of the two areas of two circles is the 80x square meters. Find the length of a radius of each circle of them is twice as

Alborosie

Answer:

$r = 4m$ --- small circle

$R =8m$ --- big circle

Step-by-step explanation:

Given

$Area = 80\pi\ m^2$ -- sum of areas

$R = 2r$

Required

The radius of the larger circle

Area is calculated as;

$Area = \pi r^2$

For the smaller circle, we have:

$A_1 = \pi r^2$

For the big, we have

$A_2 = \pi R^2$

The sum of both is:

$Area = A_1 + A_2$

$Area = \pi r^2 + \pi R^2$

Substitute: $R = 2r$

$Area = \pi r^2 + \pi (2r)^2$

$Area = \pi r^2 + \pi *4r^2$

Substitute $Area = 80\pi\ m^2$

$80\pi = \pi r^2 + \pi *4r^2$

Factorize

$80\pi = \pi[ r^2 + 4r^2]$

$80\pi = \pi[ 5r^2]$

Divide both sides by $\pi$

$80 = 5r^2$

Divide both sides by 5

$16 = r^2$

Take square roots of both sides

$4 = r$

$r = 4m$

The radius of the larger circle is:

$R = 2r$

$R =2 * 4$

$R =8m$

3 0

3 years ago