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Luda [366]
3 years ago
6

Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou

chdown
Physics
2 answers:
ch4aika [34]3 years ago
6 0

Answer:

2s

Explanation:

Given parameters:

Distance  = 30m

Speed  = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

    Speed  = \frac{disance }{time}  

 Time taken  = \frac{distance }{speed }   = \frac{30}{15}  = 2s

The time it takes is 2s

Anton [14]3 years ago
5 0

<u>Given:</u>

  • <u>30 meters</u>
  • <u>Speed of 15 m/s</u>

<u>We will solve:</u>

<u>Speed  = </u>\frac{distance}{time}<u></u>

<u></u>

<u> Time taken  = </u>\frac{distance}{time} <u>= </u>\frac{30m}{15m/s} <u>= 2s</u>

<u></u>

<u>The time it takes is 2s</u>

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Skater 1 has a mass of 45 kg and is at rest. Skater 2 has a mass of 50 kg and is moving slowly at a constant velocity of 3.2 m/s
algol13

Answer:

None

Explanation:

Force, F is given by ma where m is the mass of an object and a is acceleration

Acceleration is the rate of change in velocity per unit time. Since skaters with mass of 75 kg and 50 kg are moving at a constant speed, there is no acceleration hence F=50*0=0 and F=75*0=0

For skater of 45 kg, he is at rest to mean the initial and final velocitu of the skater is zero hence no acceleration, the force will be 45*0=0

Therefore, none of the skaters will experience a greater net force.

3 0
3 years ago
Express in words AND mathematically the relationship between period and frequency
inessss [21]
The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave.
8 0
3 years ago
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You move a 25 N object 5 meters. How much work did you do ?
olga55 [171]
The answer is 125 Joules

The first thing to take note of is the work equation: W=F×D

Since we already have our force and our distance that will help make this problem easier.

So, W=25*5

W=125

Therefore, our answer is 125 Joules since work is measured in joules

Hope this helped!! :)


3 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H
Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

3 0
3 years ago
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