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2 years ago

A projectile is shot on level ground with a

Physics

1 answer:

erastova [34]2 years ago

5 0

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above

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a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood

geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).

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3 years ago

A projectile is shot on level ground with a

erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above

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2 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

3 years ago

Why do surface winds tend to cross the isobars?

scZoUnD [109]

Geostrophic winds blows parallel to the isobars. That is because the Coriolis force and pressure gradient force ( PGF ) are in balance. But near the surface the friction can act to change the direction of the wind and to slow it down. Coriolis force decreases at the surface and PGF stays the same. The difference in terrain conditions affects how much friction is exerted. Hills and forests force the wind to change direction more than flat areas. Answer: Friction reduces the speed so Coriolis is weakened.

3 0

3 years ago

Read 2 more answers

*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa

Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
This force is not a new force, is just the net force aiming to the center of the circle.
In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
So, we can write the following expression:

$F_{cent} = F_{n} - F_{g} (1)$

It can be showed that the centripetal force is related to the speed by the following expression:
$F_{cent} = m*\frac{v^{2}}{r} (2)$

The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
Replacing (2) in (1), and solving for Fn, we get:

$F_{n} = m*\frac{v^{2} }{r} + m*g (3)$

Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

$F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)$

Replacing Fg by its value, simplifying, and solving for v, we get:

$v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)$

3 0

2 years ago