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3 years ago

A material that can be hit without shattering is? viscous, flammable, hard, malleable

1 answer:

3 0

Malleable

Malleability is a property of matter, that’s specializes in metals, in which these metals can be bended, twisted or formed into a thinner sheets, and not being able to shatter to pieces instead it can be formed into a new shape. Unlike the other three, take for instance hardness. If a hard object such as wood for example when used with an axe it breaks and it is lead to smithereens. Flammable like is when applied to fire can dramatically explode when hit it obviously breaks.

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A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/

Jlenok [28]

The magnitude and direction of the velocity of the car at t = 8.00 s is v = 7,468 m / s and the vector direction = θ = 58,984°

<h3><em>Further explanation</em></h3>

In the coordinates of the two-dimensional plane (xy) there is a position vector of a point

You can write P as:

P = xi + yi

where i denotes the unit vector in the x and y directions

While the magnitude of the vector:

$\large{\boxed{\bold{P~=~\sqrt{x^2~+~y^2}}}}$

and its direction

$\large{\boxed{\bold{\theta~=~arc~tan\frac{y}{x}}}}$

Likewise with the instantaneous speed which is a derivative of the position function of time

$V =\frac{dP}{dt}$

$V ~=~\frac{dx}{dt}i~i~+\frac{dy}{dt}j$

V = vx i + vy j

The velocity of the car as a function of time is given by v = [5.00 m / s - (0.0180 m / s²) t²] i + [2.00 m / s + (0.550 m / s²) t] j so that:

vx = 5 - 0.0180t² i

vy = 2 + 0.55t j

So the instantaneous velocity at t = 8 s becomes:

vx = 5-0.0180.8²

vx = 3,848 m / s

vy = 2+ 0.55.8

vy = 6.4

So the velocity position vector becomes:

v = 3,848 i + 6.4 y

and the magnitude becomes:

$v=\sqrt{3.848^2+6.4^2}$

v = 7,468 m / s

While the velocity vector direction:

$\theta~=~=arc~tan\frac{vy}{vx}$

$\theta~=~arc~tan~\frac{6.4}{3.848}$

θ = 58,984

<h3><em>Learn more</em></h3>

Which vector should be negative

brainly.com/question/11236351

the statements that describe a vector

brainly.com/question/4417902

the resultant vector

brainly.com/question/11194474

Keywords: velocity, vector, two-dimensional coordinates,

4 0

3 years ago

Read 2 more answers

Responsible for weather patterns on earth is what convention, radiation, conduction?

tatuchka [14]

Answer:

Convection. Convection is the transfer of heat energy in a fluid. ... As the rock's temperature rises due to conduction, heat energy is released into the atmosphere, forming a bubble of air which is warmer than the surrounding air. This bubble of air rises into the atmosphere.

Explanation:

Hope this helps!

4 0

3 years ago

A solution with a pH of 2.0 is

olya-2409 [2.1K]

a solution with a pH of 2.0 is strongly acidic

4 0

4 years ago

A tennis ball is dropped from the top of a tall building. a second tennis ball is thrown down from the same building make a stat

marusya05 [52]

Answer:

D they both are the same acceleration because they are in free fall

6 0

3 years ago

A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.

Nataliya [291]

Answer:

Approximately $1.2 \times 10^{4}\; {\rm J}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The change in the gravitational potential energy ( ${\rm GPE}$ ) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass $m$ increased by $\Delta h$ , the ${\rm GPE}$ of that object would have increased by $m\, g\, \Delta h$ .

In this question, the height of this object increased by $\Delta h = 6.0\; {\rm m}$ . The mass of this object is $m = 200\; {\rm kg}$ . Thus, the ${\rm GPE}$ of this object would have increased by:

$\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}$ .

(Note that $1\; {\rm N \cdot m} = 1\; {\rm J}$ .)

3 0

2 years ago