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3 years ago

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,

$F = ILBsin \theta$