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1 year ago

A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver

Physics

1 answer:

Akimi4 [234]1 year ago

3 0

Given:

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

To find:

Time required by ball to reach the receiver = ?

Formula used:

speed = $\frac{distance}{time}$

Solution:

The speed of the ball is given by,

speed = $\frac{distance}{time}$

Thus,

Time = $\frac{distance}{speed}$

Distance = 50 yard = 45.72 meter

Speed = 40 km/hr = 11.11 m/s

Time = 4.12 second

Hence, ball reaches the receiver in 4.12 second.

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Kayla and her friends are setting up chairs for a school play each row will contain the same number of chairs Kayla knows that t

LUCKY_DIMON [66]

Answer:

96=8*c

Explanation:

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1 year ago

A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

4 0

2 years ago

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

Where,

F = Force/Weight

A = Area

$\Delta L$ = Compression

$L_0$ = Original Length

According to the values given we have to

$\Upsilon_{steel} = 200*10^9Pa$

$\Delta L = 5.6*10^{-7}m$

$L_0 = 3.2m$

$r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2$

Replacing this values at our previous equation we have,

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

$200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}$

$F = 38272.5N$

Therefore the Weight of the object is 3.82kN

4 0

1 year ago

A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

frez [133]

(u) = 20 m/s
(v) = 0 m/s
 (t) = 4 s

0 = 20 + a(4)

4 x a = -20

so, the answer is -5 m/s^2. or -5 meter per second

8 0

1 year ago

Read 2 more answers

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

3 0

1 year ago