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3 years ago

9

A block has acceleration a when pulled by a string. if two identical blocks are glued together and pulled with twice the origina

l force, what will their acceleration be?

1 answer:

WINSTONCH [101]3 years ago

5 0

<span>F = m x a 2F = 2m x a a = 2F / 2m a = F/m The acceleration would be the same.</span>

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2..Read the following statements carefully and choose the correct option

Softa [21]

Answer:

statement 1 is correct

2 not sure

Explanation:

6 0

2 years ago

If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?

Pepsi [2]

Answer:

6 km/h

Explanation:

V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h

4 0

2 years ago

A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.

zzz [600]

Answer:

0.0257259766982 m

Explanation:

$P_2$ = Atmospheric pressure = 101325 Pa

$d_1$ = Initial diameter = 1.5 cm

$d_2$ = Final diameter

$\rho$ = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

$P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa$

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m$

The diameter of the bubble is 0.0257259766982 m

8 0

2 years ago

Which state of matter has atoms that are spread out and bouncy?

NikAS [45]

The stage where atoms are spread out and bouncy is the gas stage.

7 0

2 years ago

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa. B. 651.605 kPa. C. 448.955 kPa. D. 277.28

guajiro [1.7K]

Answer:

Option C is the correct answer.

Explanation:

Absolute pressure is sum of gauge pressure and atmospheric pressure.

That is

$P_{abs}=P_{gauge}+P_{atm}$

We have

$P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa$

Substituting

$P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa$

Option C is the correct answer.

7 0

2 years ago