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ivolga24 [154]
3 years ago
12

What is the elevation of point B? What about F?

Physics
1 answer:
kolbaska11 [484]3 years ago
7 0

Explanation:

Since I can only do this by observation, the elevation of F is approximately 850km and the elevation of B is 925km.

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The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic
Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight =   \frac{2uSin(A)}{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = \frac{2(4)Sin(60)}{10} = 0.7

ii) distance travel = Range =  R = \frac{u^{2}Sin(2A) }{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = \frac{4^{2}Sin(2*60) }{10} = 1.39

iii) Maximum Height = H = \frac{u^{2}(Sin(A))^{2}  }{2g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: \frac{4^{2}(Sin60)^{2}  }{2*10} = 0.6

4 0
3 years ago
A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a
Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged  (x)= 0.13 m

now,

weight of the mass attached = Kx

             m g = k x

             0.35 x 9.8 = k x 0.13

                k = 26.38 N/m

now, using conservation of energy

 \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2

 v = \sqrt{\dfrac{kx'^2}{m}}

 v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}

 v = \sqrt{1.6958}

    v = 1.30 m/s

6 0
3 years ago
10. A 25 kg apple cart is being pushed with an applied force of 115 N. The coefficient of friction between the ground and the ca
ziro4ka [17]

Answer:

1.1 m/s²

Explanation:

From the question,

F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ =  coefficient of friction., a = acceleration of the apple cart.

Given: F = 115 N, m = 25 kg,  μ  = 0.35

Constant: g = 10 m/s²

Substitute these values into equation 2

115-(25×10×0.35) = 25×a

115-87.5 = 25a

25a = 27.5

a = 27.5/25

a = 1.1 m/s²

8 0
3 years ago
A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0
3 years ago
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water                 ---> 1</span>

 

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

 

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m        (ANSWER)</span>

8 0
3 years ago
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