Ask question

Ask question

All categories

3 years ago

14

If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially,

the block is at rest

1 answer:

grigory [225]3 years ago

3 0

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power = F x V

$F_{H}$ = F cos0

$F_{H}$ = (30) x 4/5

$F_{H}$ = 24N

Now Calculate V using following formula

V = $V_{0}$ + at

$V_{0}$ = 0

a = $F_{H}$ / m

a = 24N / 20 kg

a = 1.2m / $S^{2}$

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = $F_{H}$ x V

Power = 24 x 4.8

Power = 115.2 W

You might be interested in

Find the derivative of y = sin(ln(5x2 − 2x))

pickupchik [31]

Answer:

$y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$

Explanation:

Let $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ and we proceed to find the derivative by the following steps:

1) $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ Given

2) $y = \sin [\ln[x\cdot (5\cdot x - 2)]]$ Distributive property

3) $y = \sin[\ln x + \ln (5\cdot x - 2 )]$ $\ln (a\cdot b) = \ln a + \ln b$

4) $y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$ $\frac{d}{dx} (\sin x) = \cos x$ / $\frac{d}{dx}(\ln x) = \frac{1}{x}$ / $\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}$ /Rule of chain/Result

3 0

3 years ago

: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the

Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}$

$V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s$

$\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}$

taking f = 0.0185

at Z₁ = Z₂

$\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}$

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P= 1.13\ psi$

c) When flow is downhill z₂-z₁ = -2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P=-0.601\ psi$

7 0

3 years ago

A girl operates a radio-controlled model car in a vacant parking lot. The girl's position is at the origin of the xy coordinate

lawyer [7]

Answer:

a) 17.20

b) 11.31

c) 14.42

d) 12.65

Explanation:

(a)

The girl is at the origin of the x,y coordinates (i.e 0,0,0 )

the position vector of the car at time 't' secs is

$\vec{r}= 2+2t^2, 6+t^3,0$

at t=2s, the position vector is

$\vec{r}= 10, 14,0$

Therefore, the the distance between the car and the girl is

$s= \sqrt{(10-0)^2+(14-0)^2+(0-0)^2)}\$

s = 17.20

(b)

The position of the car at t = 0s is $\vec{r}_0 = 2,6,0$

The position of the car at t = 2s is $\vec{r}_2 = 10,14,0$

The distance of the car traveled in the interval from t=0s to t=2 s is as follows:

$s_{02}= \sqrt{(10-2)^2+(14-6)^2+(0-0)^2)} \\ \\ s_{02} = 11.31$

(c)

The position vector of the car at time 't' secs is

$\vec{r}= 2+2t^2, 6+t^3,0$

The velocity of the car is

$\vec{v}=\dfrac{d\vec{r}}{dt}= 4t, 3t^2,0$

the direction of the car's velocity at t = 2s is going to be

$\vec{v}\mid _t=2 8, 12,0$

Thus; The speed of the car is

$v_{t=2}= \sqrt{8^2+12^2+0^2} \\ \\ v_{t=2}= 14.42$

(d) the car's acceleration is:

$\vec{a}=\frac{d\vec{v}}{dt}= 4, 6t,0$

The magnitude of car's acceleration at t=2s is

$\mid \vec{a}\mid _{t=2}=\sqrt{4^2+12^2+0^2} \\ \\ \mid \vec{a}\mid _{t=2}= 12.65$

5 0

3 years ago

Identify how slip affects the strength of a metal

Fiesta28 [93]

My answer says there is a link but not,there isn’t so here a photo of my answer.

5 0

3 years ago

How are the particles moving and and arranged in a liquid?

Ipatiy [6.2K]

Answer:

The particles in a liquid have small spaces between them, but not as small as solids. The particles in a liquid are loosely arranged which means they do not have a fixed shape like solids, but they rather take the shape of the container they are in.

6 0

3 years ago