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kiruha [24]
2 years ago
7

A stone is thrown vertically upwards with an initial velocity of 40 ms-1. Find the maximum height reached by the stone. What is

the net displacement and the total distance covered by the stone?
Physics
1 answer:
Lyrx [107]2 years ago
8 0
To find the maximum height we use the expression:
U²=2gh
Where u is the initial velocity,
g is gravitational field strength
h is the maximum height.
h=u²/2g
h=40²/20
=1600/20
=80m
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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

r = 741.6959 m

3 0
3 years ago
The answer can't be D because I got it wrong when I took my test so it has to be A,B,or C
Cerrena [4.2K]
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3 years ago
Two types of energy can be either kinetic or potential energy. Identify those two types. For each type, explain why it can be ei
love history [14]

Answer:

All forms of energy are either kinetic or potential. The energy associated with motion is called kinetic energy . The energy associated with position is called potential energy . Potential energy is not "stored energy".

Explanation:

8 0
2 years ago
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A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
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ella [17]

To calculate the mass of the fuel, we use the formula

m = V \times  \rho

Here, m is the mass of fuel, V is the volume of the fuel and its value is V =302 \ L =  302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL and  \rho is the density and its value of  0.821 g/mL.

Substituting these values in above relation, we get  

m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg

Thus, the mass of the fuel 247 .94 kg.

8 0
3 years ago
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