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2 years ago

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A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. What is the maximum height reached by the basketball from its release point?

1 answer:

Hatshy [7]2 years ago

4 0

Answer:

The last one

Explanation:

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Near the surface of the moon, the distance that an object falls is a function of time. It is given by d(t) = 2.6667t2, where t i

Eddi Din [679]

Answer:

averaage speed is v = 13 feet / s

Explanation:

The average speed is the ratio of the distance traveled in a given time interval, let's calculate the distance that the body travels in the two instants of time that give us

t = 1 s

d (1) = 2.6667 1²

d (1) = 2.6667 feet

t = 4 s

d (4) = 2.6667 4²

d (4) = 42.6672 feet

Let's calculate the speed

v = Δx / Δt

v = (42.6672 -2.6667) / (4-1)

v = 40/3

v = 13.33335 feet / s

v = 13 feet / s

6 0

2 years ago

A mechanic uses a hydraulic car jack to lift the front end of a car to change the oil. The jack used exerts 8,915 N of force fro

Vaselesa [24]

Answer:

63.05 cm²

Explanation:

We use Pascals law to find the answer.

The law says that in an incomprehensible , non-viscous fluid the pressure applied will transmit through out the fluid without a change.

So, Pressure on larger piston = pressure on smaller piston.

$\frac{444}{3.14} = \frac{8915}{A}$

A ≅ 63.05 cm²

3 0

2 years ago

01 – (Valor – 2,0) O maior campo de testes de veículos da América Latina, localizado na cidade de Indaiatuba (SP), tem forma cir

Scilla [17]

Answer:

a) Calcule a frequência em RPM

= 0.6 RPM

b) a velocidade escalar do carro em m/s.

= 20m/s

Explanation:

a) Calcule a frequência em RPM

A fórmula para calcular a frequência é: 1/T

onde T= Tempo (seconds)

T = 100s

A frequência = 1/100s

A frequência = 0.01Hz

em RPM

A fórmula para calcular a frequência em RPM =

1 Hz = 60RPM

0.01Hz =

A frequência em RPM = 0.01Hz × 60

= 0.6 RPM

b) a velocidade escalar do carro em m/s.

A fórmula para calcular a velocidade escalar = diâmetro ou distância (m) ÷ tempo (s)

Diâmetro ou Distância = 2.0km

Converter 2.0km para m

1 km = 1000m

2km =

2 km × 1000m

= 2000m

A velocidade escalar = 2000m ÷ 100s

A velocidade escalar = 20m/s

Answer:

a) Frequency in RPM

= 0.6 RPM

b) Scalar Velocity

= 20m/s

Explanation:

a) Frequently in RPM

Formula : 1/T

Where T= Time (seconds)

T = 100s

= 1/100s

= 0.01Hz

Frequency in RPM =

1 Hz = 60RPM

0.01Hz = 0.01Hz × 60

= 0.6 RPM

b) Scalar velocity

The formula = Diameter or Distance ÷ Time

Diameter or Distance = 2.0km

Convert 2.0km to m

1 km = 1000m

2km =

2 km × 1000m

= 2000m

Scalar Velocity = 2000m ÷ 100s

Scalar Velocity = 20m/s

8 0

2 years ago

If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100

SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is $8.98\times 10^{-12} C$

Explanation:

Electric field E at a point due to a point charge is given by

$E=k \frac{q}{r^2}$

where $k$ is the constant = $9.0 \times 10^9 Nm^2 / C^2$

$q$ is the magnitude of point charge and $r$ is the distance from the point charge

Charges entering one meter of axon is $5.\times 10^{11} \times (+e)$

Charges entering 0.100 mm of axon is $5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}$

substituting the value of $+e=1.6\times 10^{-19} C$ in above equation, we get charge enter a 0.100 mm length of the axon is

$q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C$

3 0

2 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

5 0

2 years ago