All categories

2 years ago

Someone help please by providing work and answers please :)

Physics

1 answer:

Nastasia [14]2 years ago

5 0

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

You might be interested in

A jogger runs by a river with a velocity of 7 m s relative to the ground. A leaf floating on the river moves with a velocity of

Hunter-Best [27]

Answer:

The velocity of the leaf relative to the jogger is 5 m/s.

Explanation:

Given that,

Velocity of jogger wrt to the ground, $V_j=7\ m/s$

velocity of leaf wrt the ground, $v_i=2\ m/s$

We need to find the velocity of the leaf relative to the jogger. Let it is equal to V. So, it is given by :

$V=v_j-v_i\\\\V=7-2\\\\V=5\ m/s$

So, the velocity of the leaf relative to the jogger is 5 m/s. Hence, this is the required solution.

3 0

3 years ago

How is power defined? Question 4 options: the quantity of work accomplished the direction of the work the total distance an obje

GarryVolchara [31]

Power can be defined as the rate at which work is accomplished.

Option D is the correct answer.

<h3></h3><h3>Power </h3>

The work done by an object in a given time interval is called the power of that object.

Suppose an external force F is applied to any object for the time interval T seconds. Due to this external force, the object will perform some amount of work for the time T seconds. This work W done by the object for the time interval T seconds is called the power of that object.

Power can be defined in mathematical term which is given below.

$P = \dfrac {W}{T} \;\rm Watts$

Thus the power can also be defined as the work done by the object per unit time interval.

Hence we can conclude that option D is the correct answer.

To know more about power, follow the link given below.

brainly.com/question/1618040.

8 0

1 year ago

Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when

nignag [31]

<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

v₁ is the final speed of car X.

$m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s$

So, car X will move with a velocity of -12 m/s.

3 0

2 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago

What are at least three types of energy involving a microwave

timama [110]

I'm pretty sure what you are trying to ask for is radiative energy, light energy, and electronic energy.
Radiative since the microwave is releasing radiation,
Light since there is light inside the microwave,
Electronic since it is plugged in and uses electricity.
You can also use sound, but I don't think every microwave makes sound.

8 0

3 years ago