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3 years ago

12

Over a period of 2.50 seconds, a speedboat accelerates uniformly from 18.5 m/s to 46.5 m/s. What is the acceleration of the spee

dboat relative to the shore?

1 answer:

pashok25 [27]3 years ago

7 0

The answer is : 11.2 m/s2

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In order to be considered scientific, evidence must be:

zhenek [66]

Reviewed by many sources. for example, it must be reviewed by scientist too

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4 years ago

An Airbus A320 jetliner has a takeoff mass of 75,000 kg. It reaches its takeoff speed of 82 m/s (180 mph) in 35 s. What is the t

timofeeve [1]

Answer:

175.5 kN

Explanation:

We find the acceleration of the Airbus A320 jetliner from,

a = (v - u)/t where u = initial velocity of jetliner = 0 m/s (since it starts from rest), v = final velocity of jetliner = 82 m/s and t = time for velocity change = 35 s

So a = (v - u)/t = (82 m/s - 0 m/s)/35 s = 82 m/s ÷ 35 s = 2.34 m/s²

Now, the thrust of the engines on the jetliner T = ma where m = mass of jetliner = 75,000 kg and a = acceleration of jetliner = 2.34 m/s²

T = ma

= 75,000 kg × 2.34 m/s²

= 175500 N

= 175.5 kN

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3 years ago

Stated that animals,as well as plants, are made of cells

koban [17]

Yes everything is made of cells because without cells our being can't stay together.

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4 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

4 years ago

Can anybody please help? 15 points pls

Mariana [72]

Acceleration= v/ r
So you have to divide 700 by 0.800 which is equal to 875 m/s 2

5 0

3 years ago