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2 years ago

Any one watch SerieS here

Physics

2 answers:

Illusion [34]2 years ago

8 0

Answer:

nope

Explanation:

never even heard of it

levacccp [35]2 years ago

5 0

Answer:

no what is about if u dont mind explaining it to me?

UwU

If not have a good no a great day anyway

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A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0

3 years ago

Which BEST describes the difference between speed and velocity?

Murljashka [212]

D. velocity includes rate of change and direction

8 0

3 years ago

Read 2 more answers

Name the 2 types of tissue that form your skin?

QveST [7]

Answer:

Epithelial tissue and Muscle tissue

Explanation:

7 0

2 years ago

Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,

SVEN [57.7K]

Answer:

Buffalo, NY

Explanation:

Temperature in Buffalo, NY = -29°C

In order to compare the temperatures we need to convert them to the same scale.

$F=\dfrac{9}{5}C+32\\\Rightarrow F=\dfrac{9}{5}\times -29+32\\\Rightarrow F=-20.2$

So, the temperature in Buffalo, NY was -20.2°F and the temperature in Anchorage, AL was 19°F.

Hence, it was colder in Buffalo, NY than in Anchorage, AL.

3 0

2 years ago

20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

aliina [53]

$Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2} \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot \frac{4m^3}{6.5m^3} \cdot \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C$

7 0

3 years ago