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Fiesta28 [93]
3 years ago
11

If we assume that, in the African savanna, rainfall and the amount of grass is plentiful, what other factor would BEST explain a

reduced antelope population?
Physics
2 answers:
valentina_108 [34]3 years ago
8 0
Diseases, Hunting (sport for humans),  predators
777dan777 [17]3 years ago
3 0

Increased number of predators

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A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed <img src="https:
s344n2d4d5 [400]

Answer:

a) v_com= Rω

b) -2.254 m/s^{2}

c) 51.2 rad/s^{2}

d) t=1.08 seconds

e) x=7.865m

f) v_roll=6.07m

Explanation:

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

\alpha=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

a) Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

b)    F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

c) α=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

d) t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

e) X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

f) v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

7 0
3 years ago
An ___ wave carries energy through matter
garik1379 [7]
A sound wave is the answer
4 0
3 years ago
What is the acceleration of a 5 kg mass pushed by a 10 n force?
pav-90 [236]
<span>2Kg50m/
s2.5m/
s2<span>2m/
s2</span></span>
6 0
2 years ago
Read 2 more answers
A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
pantera1 [17]

Answer:

Part a)

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

v = \sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{L}

now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

time taken by the wave to reach the top is given as

t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}

t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})

t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})

t = 12 s

4 0
3 years ago
How much kinetic energy does a 1500 kg car have if it’s moving at 15 m/s
Tresset [83]

Answer:

Mass of car m = 1500 kg

velocity v = 15 m/s

kinetic energy = ½ mv2

= ½ x 1500 x (15)2    

= 1687501   kg m2 /s2

  = 168750J

Explanation:

6 0
3 years ago
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