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Marrrta [24]
3 years ago
11

The place kicker on a football team kicks a ball from ground level with an initial speed of 5.80 m/s at an angle of 17.0° above

the horizontal. How long is the ball in the air before it lands on the ground again
Physics
1 answer:
Likurg_2 [28]3 years ago
8 0

Answer:

0.34 s

Explanation:

Given that,

Initial speed of a ball, u = 5.8 m/s

It is kicked at an angle of 17.0° above the horizontal.

The vertical component of velocity will be,

u_y=u\sin\theta\\\\=5.8\times \sin 17\\\\=1.69\ m/s

Let it takes t time in the air before it lands on the ground again. It can calculated as :

t=\dfrac{2u_y}{g}\\\\t=\dfrac{2\times 1.69}{9.8}\\\\t=0.34\ s

So, it will take 0.34 seconds.

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Question 8 (4 points)
katrin2010 [14]

Answer:

The answer to your question is:

Explanation:

Data

Duane                                          Albert

d = 5 m ;  v = 3 m/s                     v = 4.2 m/s

a)                                                b)

Duane's                                      Albert's

d = 5 + (3)t                                  d = 4.2t

d = 5 + 3t

c)                            5 + 3t = 4.2t

                              4.2t - 3t = 5

                                      1.2t = 5

                                           t = 4.17 s

d)

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

4 0
3 years ago
Read 2 more answers
A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d
pentagon [3]

Answer:

The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}

Now, plug in the given values and solve for 'a'. This gives,

a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}

Now, plug in the given values and solve for 'S'. This gives,

S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0
3 years ago
A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
3 years ago
When did the ball fly farther—when it hit the moving bat or the bat that did not move?
8_murik_8 [283]

Answer:

when it hit the moving bat

Explanation:

force equals mass times acceleration which means the moving bat will add more force to the ball.

6 0
3 years ago
What is the resistance of the coil A at 600 kelvin if its resistance at 300 kelvin is 50 ohms? (Assume the temperature coefficie
leonid [27]

155Ω

Explanation:

R = R ref ( 1 + ∝ ( T - Tref)  

where R = conduction resistance at temperature T

R ref = conductor resistance at reference temperature

∝ = temperature coefficient of resistance for conductor

T = conduction temperature in degrees Celsius

T ref = reference temperature that ∝ is specified at for the conductor material

T = 600 k - 273 k = 327 °C

Tref = 300 - 273 K = 27 °C

R = 50 Ω ( 1 + 0.007 ( 327 - 27) )

R = 155Ω

8 0
3 years ago
Read 2 more answers
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