Ask question

Ask question

All categories

3 years ago

7

PLEASE HELP !!! I NEED TO PASS

1 answer:

zhannawk [14.2K]3 years ago

7 0

Answer:

Sorry I don't know the answer .Hope other help you.sorryyyyyyyy

You might be interested in

A 1400-kg car moving with a speed of 32 m/s is approaching a stoplight. The light turns yellow and the car makes an abrupt stop

inna [77]

Answer:

c. 716, 800 J

Explanation:

t = Time taken

u = Initial velocity = 32 m/s

v = Final velocity = 0

s = Displacement = 60 m

a = Acceleration

m = Mass of car = 1400 kg

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-32^2}{2\times 60}$

Work done is given by

$W=Fscos\theta\\\Rightarrow W=mascos\theta\\\Rightarrow W=1400\times \dfrac{0^2-32^2}{2\times 60}\times 60\times cos0\\\Rightarrow W=-716800\ J$

The amount of work done to stop the car is 716800 J

8 0

3 years ago

A particle of unit mass moves in a potential V(x)= ax^2+b/x^2 . Where a and b are constnts. Calcuate the angular frequency of sm

Sholpan [36]

Answer:

Explanation:

Given

Potential Energy is given by

$U(x)=ax^2+\frac{b}{x^2}$

And Force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

Particle will be at equilibrium when Potential Energy is either minimum or maximum

$F=-\left ( 2ax-\frac{2b}{x^3}\right )$

i.e. $ax=\frac{b}{x^3}$

$x_0=(\frac{b}{a})^{0.25}$

So angular Frequency of small oscillation is given by

$\omega =\sqrt{\frac{U''(x)}{m}}$

for $m=1$

we get $\omega =\sqrt{\frac{U''(x_0)}{1}}$

$U''(x_0)=2a+6a= 8a$

$\omega =\sqrt{8a}$

4 0

3 years ago

Gold has a density of 19.3 g/cm3. What is the mass of a 5 cm3 block of gold?

Tems11 [23]

Do not forget that mass = <span>volume x density
</span>Mass of 1 cm^3 = Density[/tex]
$mass of 2 cm^3 = 19.3 g + 19.3 g = 2*19.3 g$
Then eventually we can find <span>mass of 5 cm^3 : =
</span> $19.3 g + 19.3 g+19.3g+19.3g+19.3g= 5*19.3 g$
So the answer is D
<span>And that's it. I'm sure it will help.</span>

5 0

3 years ago

A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n

RSB [31]

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

3 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago