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o-na [289]
3 years ago
11

if an object is thrown straight up with an initial velocity of 8m/s and takes 3 seconds to strike the ground, from what height w

as the object thrown?​
Physics
1 answer:
KatRina [158]3 years ago
7 0

Answer:

s = it+1/2 at²

s= 8×3+1/2 (10)(3)²

s = 24+45

s= 69

the object was thrown from a height of 69 meters

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what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s
kotegsom [21]
Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.
7 0
3 years ago
if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before att
omeli [17]

pavement is defined as the surface of Road or sidewalk.

for example, the surface of Expressway.

There are two types of pavement.

rigid pavement which consists of one layer.

flexible pavement which consist of multiple layers.

While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the pedal, then apply brake lightly until we are moving at a low speed.

if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before attempting to return to the pavement steer straight ahead and speed up apply the brakes hard

To know more about pavement:

brainly.com/question/28456065

#SPJ4

5 0
2 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
students may elect to take the Knowledge Test online or at Drivers Office what additional test must be taken at the licensing of
Nutka1998 [239]
None I mean you have to take a drivers test but you will have to take the knowldege test twice

4 0
3 years ago
The nuclear equation is incomplete.<br><br><br>What particle completes the equation?​
Romashka-Z-Leto [24]

Answer:

_{6}^{12} C

Explanation:

_{1}^{1}H+_{7}^{15}N --->    _{6}^{12} C+ _{2}^{4}He

1+7 = 6+2 =8 -protons

1+15 = 12+4 = 16 -  protons +neutrons

3 0
4 years ago
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