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3 years ago

Calculate the average (mean) of these numbers: 6, 5, 7, 9.8*

Physics

1 answer:

irina1246 [14]3 years ago

5 0

Answer:

13.94 to 2 d.p

Explanation:

just add them all up and i am confused the ones you said or the ones listed ?

the one i did was the listed ones

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A system can only achieve a lower energy state by:_______

bezimeni [28]

Transferring its energy to its surroundings.

7 0

2 years ago

A jet airplane has a velocity of 1145 knots. A knot is 1 nautical mile (nm)/hr. A nautical

Tanya [424]

Answer:

1 m = 39.37 in = 39.37/12 ft = 3.28 ft

V = 1145 k/hr = 1145k/hr * 6076 ft/k = 6957020 ft / hr

V = 6957020 ft/hr / 3600 s/hr = 1933 ft/sec

V = 1933 ft/sec / (3.28 ft / m) = 589 m/s

Check:

88 ft/sec = 60 mph

(1145 k/hr * 6076 ft / k) 3600 sec/hr = 1933 ft/sec = 589 m/s

1933 ft/sec / (88 ft/sec) * 60 mph = 1318 mph

Also, 1318 / 1145 = 6076 / 5280 as it should

4 0

2 years ago

P.E

Vladimir79 [104]

Answer:

pagtalon?

Explanation:

because that had to be the answer

5 0

3 years ago

A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the

Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$ . What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

$= 3050.19 \ kg \ m^2$

So the moment of inertia is $3050.19 \ kg \ m^2$ .

4 0

2 years ago

problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi

Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

Explanation:

(a) To reach maximum distance

$g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\ x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m$

(b) For Time

To find t we must find t1 and t2

t=t1+t2

For T1

$t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s$

For T2

$x_{l}=Vbt+(1/2)gt_{2}^{2}\\ as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s$

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0

3 years ago