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2 years ago

12

A cat leaps to catch a bird. if the cat's jump was at 60.0° off the ground and its initial velocity was 2.74 m/s, what is the hi

ghest point of its trajectory?

1 answer:

user100 [1]2 years ago

7 0

Answer:

0.29 m

Explanation:

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A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

6 0

2 years ago

S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

2 years ago

What is angular velocity

butalik [34]

Answer:

angular velocity(ω) is the rate change of angular displacement.

ω=θ/t and it SI unit is rad/s

Explanation:

this is very similar with the definition of linear velocity (rate of change of displacement). it specifies the angular speed of an object and the axis about which the object is rotating.

7 0

2 years ago

If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?

aleksklad [387]

Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right

6 0

2 years ago

An object releasing energy is evidence of which type of chemical change?

Ray Of Light [21]

Exothermic is the answer to your question

8 0

2 years ago