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2 years ago

It is easier to climb up a slanted slope than a vertical slope

2 answers:

4 0

The correct answer is it is easier to climb a slanted slope rather then a vertical slope. Vertical is straight up versus a nice slant.

V125BC [204]2 years ago

3 0

IT IS EASIER TO CLIMB A SLANTED SLOPE

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A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast

Neko [114]

Answer:

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of $\frac{1}{12}$ × 360 = 30m

And turtle covers a distance of $\frac{11}{12}$ × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = $\frac{30m}{2400s}$ = 0.0125m/s

Turtle's speed = $\frac{330m}{2400s}$ = 0.1375m/s

8 0

2 years ago

Read 2 more answers

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one hold 163.296 cm³ of paint

5 0

2 years ago

A rectangular beam 10 cm wide, is subjected to a maximum shear force of 50000 N, the corresponding maximum shear stress being 3

nika2105 [10]

Answer:

Option B is the correct answer.

Explanation:

Shear stress is the ratio of shear force to area.

We have

Shear stress = 3 N/mm² = 3 x 10⁶ N/m²

Area = Area of rectangle = 10 x 10⁻² x d = 0.1d

Shear force = 50000 N

Substituting

$\texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm$

Width of beam = 16.67 cm

Option B is the correct answer.

6 0

2 years ago

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

6 0

2 years ago

A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(

maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

7 0

2 years ago