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Murljashka [212]
3 years ago
11

Ian walks 2 km to his best friends house, then walks 0.5 km to the library. He then makes 2.5 km walk home. The entire walk took

him 3 hours. What is his average speed in km/h?
SHOW YOUR WORK PLEASE



2. A truck speeds up from a velocity of 6 m/s to 14 m/s in 4 seconds. What is the trucks acceleration? SHOW YOUR WORK



3. A car starts from rest and reaches 15 m/s in 2.5 seconds. What is the acceleration of he car? SHOW YOUR WORK
Physics
1 answer:
allochka39001 [22]3 years ago
5 0
1. 2+0.5+2.5= 3. 2km/hr average




2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration


3. 15m/s divided by 2.5 = 6m/s acceleration
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mel-nik [20]

Answer:

Explanation:

m = ρV = 1.03( 1000 kg/m³)(π(2² m²)(3.0 m)) = 12360π kg

m ≈ 38,830 kg

5 0
3 years ago
"determine the resultant internal loadings acting at the cross sections at points f and g of the frame. set θ = 27º and t = 178
Leviafan [203]

Hi you didn't provide any images to solve the question, hence I am going to solve a different question of same concept so you can have an idea how to tackle such types of questions.(please refer to the attachment for question)

Answer:

<u> Please refer to the attachment for answers and explanation</u>

Explanation:

<u> Please refer to the attachment for answers and explanation</u>

5 0
3 years ago
What mass of steam at 100°C must be mixed with 119 g of ice at its melting point, in a thermally insulated container, to produce
N76 [4]

Answer:M=27.92\ gm

Explanation:

Given

mass of ice m=119\ gm

Final temperature of liquid T_f=57^{\circ}C

Specific heat of water c=4186\ J/kg-K

Latent heat of fusion L=333\ kJ/kg

Latent heat of vaporization L_v=2256\ kJ/kg

Suppose M is the mass of steam at 100^{\circ} C

Heat required to melt ice and convert it to water at 57^{\circ}C

Q_1=mL+mc(T_f-0)

Heat released by steam

Q_2=ML_v+Mc(100-T_f)

Q_1 and Q_2 must be equal as the heat gained by ice is equal to Heat released by steam

Q_1=Q_2

\Rightarrow mL+mc(T_f-0)= ML_v+Mc(100-T_f)

\Rightarrow M=\dfrac{m[L+c\times T_f]}{L_v+c(100-T_f)}

\Rightarrow M=\dfrac{119[333\times 10^3+4186\times 57]}{2256\times 10^3+4186\times (100-57)}

\Rightarrow M=119\times 0.2346

M=27.92\ gm

7 0
3 years ago
How much work does it take to accelerate a 9.0 kg object from rest to 27 m/s?
adoni [48]
Initial kinetic energy = 0J (v=0 (rest))

Final kinetic energy = 1/2mv ²
Ek=1/2(9)(27 ²)
Ek=4.5(729)
Ek=3280.5


∆Energy=3280.5J
(As it starts from 0)

Work= ∆energy
So work =3280.5J

Answer=3300J (forth option)
5 0
3 years ago
It’s the 18th century and you are responsible for artillery. Victory hangs in the balance and it all depends on you making a goo
sweet [91]

Answer:

a) You should position the cannon at 981 m from the wall.

b) You could position the cannon either at 975 m or 7.8 m (not recomended).

Explanation:

Please see the attached figure for a graphical description of the problem.

In a parabolic motion, the position of the flying object is given by the vector position:

r =( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

where:

r = position vector

x0 = initial horizontal position

v0 = module of the initial velocity vector

α = angle of lanching

y0 = initial vertical position

t = time

g = gravity acceleration (-9.8 m/s²)

The vector "r" can be expressed as a sum of vectors:

r = rx + ry

where

rx = ( x0 + v0 t cos α ; 0)

ry = (0 ; y0 + v0 t sin α + 1/2 g t²)

rx and ry are the x-component and the y-component of "r" respectively (see figure).

a) We have to find the module of r1 in the figure. Note that the y-component of r1 is null.

r1 = ( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

Knowing the the y-component is 0, we can obtain the time of flight of the cannon ball.

0 = y0 + v0 t sin α + 1/2 g t²

If the origin of the reference system is located where the cannon is, the y0 and x0 = 0.

0 = v0 t sin α + 1/2 g t²

0 = t (v0 sin α + 1/2 g t)         (we discard the solution t = 0)    

0 = v0 sin α + 1/2 g t

t = -2v0 sin α / g

t = -2 * 100 m/s * sin 53° / (-9.8 m/s²) = 16.3 s  

Now, we can obtain the x-component of r1 and its module will be the distance from the wall at which the cannon sould be placed:

x = x0 + v0 t cos α

x = 0 m + 100m/s * 16.3 s * cos 53

x = 981 m

The vector r1 can be written as:

r1 = (981 m ; 0)

The module of r1 will be: x = \sqrt{(981 m)^{2} + (0 m)^{2}}

<u>Then, the cannon should be placed 981 m from the wall.</u>

b) The procedure is the same as in point a) only that now the y-component of the vector r2 ( see figure) is not null:

r2y = (0 ; y0 + v0 t sin α + 1/2 g t² )

The module of this vector is 10 m, then, we can obtain the time and with that time we can calculate at which distance the cannon should be placed as in point a).

module of r2y = 10 m

10 m = v0 t sin α + 1/2 g t²

0 = 1/2 g t² + v0 t sin α - 10 m

Let´s replace with the data:

0 = 1/2 (-9.8 m/s² ) t² + 100 m/s * sin 53 * t - 10 m

0= -4.9 m/s² * t² + 79.9 m/s * t - 10 m

Solving the quadratic equation we obtain two values of "t"

t = 0.13 s and t = 16.2 s

Now, we can calculate the module of the vector r2x at each time:

r2x = ( x0 + v0 t cos α ; 0)

r2x = (0 m + 100m/s * 16.2 s * cos 53 ; 0)

r2x = (975 m; 0)

Module of r2x = 975 m

at t = 0.13 s

r2x = ( 0 m + 100m/s * 0.13 s * cos 53 ; 0)

r2x = (7.8 m ; 0)

module r2x = 7.8 m

You can place the cannon either at 975 m or at 7.8 m (see the red trajectory in the figure) although it could be dangerous to place it too close to the enemy fortress!

7 0
3 years ago
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