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3 years ago

14

Which player is usually the best ball-handler on the court?

2 answers:

Alekssandra [29.7K]3 years ago

7 0

I would think after watching 3 basket ball games the point gaurd.

siniylev [52]3 years ago

6 0

I think it's D)!!!!!!!!!!

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What is power the quotient of?

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Power is equal to work divided by time

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3 years ago

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Britain and France signed an entente and became the

V125BC [204]

<span> Allied Forces. they became the allies.</span>

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4 years ago

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What is the volume of an object that has a mass of 675 g and a density of 15 g/cm3 ?

SpyIntel [72]

Answer:

45

Explanation:

because you divide volume and mass

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3 years ago

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car

Alex73 [517]

Answer:

15.8640053791 s

392.780107582 m

29.5184032275 m/s

Explanation:

0 denotes initial

x denotes displacement

c denotes car

t denotes truck

r denotes rear

$x_0_{cr}=-49.5\ m$

$a_c=0.6\ m/s^2$

$x_0_{t}=0$

$v_0_{c}=v_0_{t}$

For the car

$x_c=x_0+v_0_{c}t+\dfrac{1}{2}at^2$

The displacement of the truck will be

$x_t=v_tt$

From the above two equations we get

$x_c-x_t=x_0+v_0_{c}+\dfrac{1}{2}at-v_{t}t=26\\\Rightarrow 26=x_0+\dfrac{1}{2}at^2\\\Rightarrow 26+49.5=\dfrac{1}{2}0.6t^2\\\Rightarrow t=\sqrt{\dfrac{2(26+49.5)}{0.6}}\\\Rightarrow t=15.8640053791\ s$

The time taken is 15.8640053791 s

$x-x_0=v_{0}_{c}t+\dfrac{1}{2}at^2\\\Rightarrow x-x_0=20\times 15.8640053791+\dfrac{1}{2}0.6\times 15.8640053791^2\\\Rightarrow x-x_0=392.780107582\ m$

The distance the car travels is 392.780107582 m

$v=v_0+at\\\Rightarrow v=20+0.6\times 15.8640053791\\\Rightarrow v=29.5184032275\ m/s$

The velocity of the car is 29.5184032275 m/s

6 0

3 years ago

A0.350 kg iron horseshoe that is initially at 600°C is dropped into a bucket containing 21.9 kg of water at 21.8°C. What is the

Svetach [21]

Answer: $22.8^0C$

Explanation:-

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$

where,

$m_1$ = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]

$m_2$ = mass of water = 21.9 kg = 21900 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of iron horseshoe = $600^oC$

$T_2$ = temperature of water = $21.8^oC$

$c_1$ = specific heat of iron horseshoe = $0.450J/g^0C$

$c_2$ = specific heat of water = $4.184J/g^0C$

Now put all the given values in equation (1), we get

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$

$350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)]$

$T_{final}=22.8^0C$

Therefore, the final equilibrium temperature is $22.8^0C$ .

3 0

3 years ago