Ask question

Ask question

All categories

schepotkina [342]

3 years ago

6

3. A particle is projected to the right from the position S = 0, when an initial velocity of 8 m/s. If the acceleration of the p

article is defined by the relation a = -0.5 v3/2, where a in m/s2 and v in m/s. Determine a) the distance the particle will have traveled when its velocity is 5 m/s b) the time when v = 1m/s c) the time require for the particle to travel 8m

1 answer:

Alenkasestr [34]3 years ago

6 0

Answer:

a) 3.5 m

b) 14 secs

c) 1.4 secs

Explanation:

<u>a) Determine the distance the particle will travel</u>

given velocity ( final velocity ) = 5 m/s

v^2 = u^2 + 2as

s = ( v^2 - u^2 ) / 2a

= ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )

= 3.5 m

<u>b) Determine the time when v = 1m/s</u>

V = u + at

1 = 8 + ( -0.5 * 1^3/2 ) * t

∴ t = 14 secs

c) Determine the time required for particle to travel 8 m

<em>we will employ both equations above </em>

V^2 = u^2 + 2as

s = 8 m , V = unknown , u = 8 m/s back to equation

V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8

∴ V^2 + 8V^3/2 - 64 = 0

resolving the above equation

V = 3.478 m/s

now using the second equation

V = u + at

3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t

hence : t = 1.4 secs

You might be interested in

A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is

Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = $\frac{A}{A-C}$

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = $\frac{A}{A-C}$

= $\frac{1034}{1034-675 \cdot 6}$

Apparent specific gravity = 2.88

b) Bulk specific gravity $G_{B}^{O D}=\frac{A}{B-C}$

$G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}$

= 2.76

c) Bulk specific gravity (SSD):

$G_{B}^{S S D}=\frac{B}{B-C}$

$=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}$

$G_{B}^{S S D}$ = 2.80

d) Absorption% :

$=\frac{B-A}{A} \times 100 \%$

$=\frac{1048 \cdot 9-1034}{1034} \times 100$

Absorption = 1.44 %

e) Bulk Volume :

$v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}$

$=\frac{1048 \cdot 9-675 \cdot 6}{1}$

= $373.3 cc$

5 0

2 years ago

A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

3 0

2 years ago

Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

WARRIOR [948]

Answer:D. Location, vehicle type, and driving habits

5 0

2 years ago

Read 2 more answers

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago

A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD

Stels [109]

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

900

=33.77

5 0

3 years ago

Read 2 more answers