An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (A) What was the takeoff speed?

1 answer:

7 0

Answer:

9.6 m/s

Explanation:

Angle of projection, θ = 28°

Horizontal distance, R = 7.8 m

Let the velocity of projection is given by u.

The formula used to find the velocity of projection is given by

$R=\frac{u^{2}Sin2\theta }{g}$

$7.8=\frac{u^{2}Sin\left (2\times 28 \right ) }{9.8}$

$7.8=\frac{u^{2}\times 0.829}{9.8}$

u = 9.6 m/s

Thus, the velocity of projection is 9.6 m/s.

You might be interested in

timurjin [86]

You have done something incorrectly. such as your data.

5 0

2 years ago

Zarrin [17]

Answer:

It's speed is 48 km/hr.

Explanation:

6 0

2 years ago

vagabundo [1.1K]

Answer:

Q = 1.2*10⁻¹² C

Explanation:

For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

$C = \frac{Q}{V} (1)$

For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

$C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)$

So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

$Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)$