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Ronch [10]
3 years ago
8

How long does a player sit out of a game of handball for committing a second or third foul?

Physics
1 answer:
yulyashka [42]3 years ago
5 0

Answer:

Walking’ - If a handball player takes more than three steps without dribbling (bouncing the ball) or holds the ball for more than 3 seconds without bouncing it, shooting or passing, then that is deemed ‘walking' and possession is lost.

'Double dribble’ - Handball players cannot receive the ball and bounce it, then hold the ball, and bounce it again. This is termed ‘double dribble’ and is against the rules.

Askmeanything2♡

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Rodrigo wants to find more ways to improve his latest design. Which of the following should he try?
irina [24]

Answer: ask other people if they like it, or ask what they want to add

Explanation: because maybe other people can can help him improve and brainstorm with other’s.

5 0
3 years ago
Read 2 more answers
A boy is holding a ball 1 m from the ground with a force of 20 N. He holds it still for 60seconds. How much power in watts is be
Sergio [31]

Answer:

Power = 0.33 Watts

Explanation:

Given the following data;

Distance = 1m

Force = 20N

First of all, we would solve for the work done by the boy.

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 20*1 = 20J

Now to find power;

Power = workdone/time

Power = 20/60

Power = 0.33 Watts.

8 0
3 years ago
The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.
scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347

Dividing the first two equations we get

1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0
3 years ago
How many moles of nitrogen are needed to completely convert 6.34 mol of hydrogen?
____ [38]
If we use the equation:
N2 + 3H2 --> 2NH3
Then
1 mol of Nitrogen required 3 moles of Hydrogen
x mols : 6.34mols
X = 6.34/3
X = 2.11 moles of Nitrogen are required.
4 0
3 years ago
Read 2 more answers
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r
dusya [7]

Answer:

28 degree C

Explanation:

We are given that

T_1=4.00^{\circ}

R_1=217.7 \Ohm

R_2=215.1\Ohm

\alpha=-5.00\times 10^{-4}C^{-1}

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

\alpha(T_2-T_1)=\frac{R_2}{R_2}-1

Using the formula

-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1

-5\times 10^{-4}(T_2-4)=0.988-1=-0.012

T_2-4=\frac{0.012}{5\times 10^{-4}}=24

T_2=24+4=28^{\circ}C

Hence, the temperature  on a spring day 28 degree C.

7 0
3 years ago
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