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Sauron [17]
3 years ago
11

Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?

Physics
1 answer:
musickatia [10]3 years ago
8 0

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

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Assume that a resistor is connected between the 150 V terminal and the common terminal. The voltmeter is then connected to an un
Lena [83]

Answer: 316.8V

Explanation:

given data:

metre moving current = 0.96mA

meters voltage = 288V

or  0.96*300V = 288V

<u><em>Solution:</em></u>

<u><em /></u>v1 = (0.96mA*150)<u><em /></u>

<u><em /></u>= 144V<u><em /></u>

<u><em /></u>

i1 = \frac{144v}{750}

= 0.192mA

i2 = imovement + i1

i2= 0.96mA+0.192mA

= 1.152mA

Vmeasured = 144V+(150)(1.152mA)

=316.8V

the unknown voltage is 316.8V

7 0
3 years ago
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
A satellite orbits the earth at a speed of 8.0x 10^3m/s . Calculate the period of orbit of the satellite if the distance between
gladu [14]

Hi there!

The period of an orbit can be found by:

T = \frac{2\pi r}{v}

T = Period (? s)
r = radius of orbit (6400000 m)

v = speed of the satellite (8000 m/s)

This is the same as the distance = vt equation. The total distance traveled by the satellite is the circumference of its circular orbit.

Let's plug in what we know and solve.

T = \frac{2\pi (6400000)}{8000} = \boxed{5026.55 s}

8 0
1 year ago
A constant force of 11.8 N in the positive x direction acts on a 4.7-kg object as it moves from the origin to the point (1.6i –
zhenek [66]

Answer:

W = 18.88 J

Explanation:

Given that,

Constant force, F = 11.8 N (in +x direction)

Mass of an object, m = 4.7 kg

The object moves from the origin to the point (1.6i – 4.6j) m

We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :

W=F{\cdot} d\\\\W=(11.8i){\cdot} (1.6i-4.6j)\\\\=11.8\times 1.6\\\\=18.88\ J

So, the work done by the given force is 18.88 J.

5 0
3 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
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