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2 years ago

If a car is traveling along a highway at a speed of 30 meters/second, what distance will it cover in one minute ?

2 answers:

5 0

The answer would be E, 1800. Why? bc if you're going 30 meters per second and trying to figure out how far you would go in a minute, you would multiply 30 x 60 bc there's 60 seconds in a minute. Giving you 1800 meters. hope this helped, have an amazing day :)

TiliK225 [7]2 years ago

5 0

The correct answer would be
E) 1,800 meters.

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A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

frez [133]

(u) = 20 m/s
(v) = 0 m/s
 (t) = 4 s

0 = 20 + a(4)

4 x a = -20

so, the answer is -5 m/s^2. or -5 meter per second

8 0

2 years ago

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

evablogger [386]

Answer:

$\text{heat loss} = 24864.05 \ W/m^2$

Explanation:

$T_1$ , $T_2$ are temperatures of gasses and liquid in Kelvins,

$t_1$ and $t_2$ are thicknesses of gas layer and steel slab in meters,
$h_1$ , $h_2$ are convection coefficients gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ is the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ are thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

using known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Using the rate equation :

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Similarly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature distribution is shown in the attached image

3 0

2 years ago

Small frogs that are good jumpers are capable of remarkable accelerations. One species reaches a takeoff speed of3.7 m/s in60 ms

MAVERICK [17]

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = $6 *10^{-3} seconds$

So acceleration of frog = $\frac{3.7}{60*10^{-3}}$

= 61.66 m/ $s^2$

So acceleration of frog is 61.66 m/ $s^2$

o it is evident that frog is capable of remarkable accelerations.

8 0

2 years ago

Read 2 more answers

Suppose that, instead of the Coulomb force law, one finds experimentally that the force between any two charge q1 and q2 is Writ

denpristay [2]

Answer: E= KQ/r^2

Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.

The magnitude of an electric field E, at a point is given by Coulombs law as

E/ F/q

Where F= Coulombs force exertedon the charge and q= electric charge

E= F/q=(KQq)/r^2q

E=KQ/r^2

6 0

2 years ago

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

weqwewe [10]

Answer:

a) v = 1.19 m / s , b) P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

Q = A v

The area of a circle is

A = π r2 = π d²/4

v = Q / A = Q 4 / pi d²

v = 0.006 4/π 0.08²

v = 1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0

Rho (water) = 1000 kg / m³

P₁ + rho y₁ = P₂ + ½ rho v₂²

P₁ = P₂ + ½ rho v₂² - rho g y₁

P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

P₁ = 1.013 10⁵ +708.5 - 9800

P₁ = 92208.5Pa

P₁ = 0.922 10⁵ Pa

8 0

2 years ago