Ask question

Ask question

All categories

4 years ago

5

A water skier is pulled behind a boat. When the skis push down on the water, what is the reaction force?

2 answers:

IrinaVladis [17]4 years ago

8 0

The answer is C) The water pushed up on the skis

The water reacts to the downward force of the skis by pushing back up against the skis.

statuscvo [17]4 years ago

8 0

Answer:

the water pushes up on the skis

Explanation:

You might be interested in

Sound waves travel faster in : Group of answer choices

ira [324]

Answer:

its a solid it travals fasters in a solid

Explanation:

6 0

3 years ago

Middle C also known as "do" on the fixed do-solfege scale, is a note used when playing or singing music. The frequency of middle

saveliy_v [14]

Answer:

198.2m/s

Explanation:

Speed of a wave(v) is the product of the frequency of the wave (f) and its wavelength(¶).

Mathematically, v = f/¶

Given frequency of the middle C = 261.63Hz

Wavelength = 131.87cm

Converting this to meters we have;

131.87/100 = 1.32m

Speed of the sound = 261.63/1.32

Speed of the sound = 198.20m/s

Therefore the speed of sound for middle C is 198.2m/s

8 0

4 years ago

A vertical piston-cylinder device initially contains 0.1 m^3 of air at 400 K and 100 kPa. At this initial condition, the piston

jenyasd209 [6]

Answer:

$Q=-38.15kJ$

Explanation:

From the question we are told that

Piston-cylinder initial Volume of air $v_1=0.1 m^3$

Piston-cylinder initial temperature $T_1=400k$

Piston-cylinder initial pressure $P_1= 100kpa$

Supply line temperature $T_s=400k$

Supply line pressure $P_s= 500kpa$

Valve final pressure $P_v=500kpa$

Piston movement pressure $P_m=200kpa$

Piston-cylinder final Volume of air $v_2=0.2 m^3$

Piston-cylinder final temperature $T_2=440k$

Piston-cylinder final pressure $P_2= 500kpa$

Generally the equation for conservation of mass is mathematically given by

$Q=m_2 \mu_2-m_1 \mu_1 +W-(m_2-m_1)h$

where

Initial moment

$m_1=\frac{p_1 V_1}{RT_1}$

$m_1=\frac{100*0.1}{0.287*400}$

$m_1=8.7*10^-^2kg$

Final moment

$m_2=\frac{p_2 V_2}{RT_2}$

$m_1=\frac{500*0.3}{0.287*440}$

$m_1=79*10^{-2}kg$

Work done by Piston movement pressure

$W=Pd$

$W=P(v_2-v_1)$

$W=200(0.2-0.1))$

$W=20000J$

Heat function

$h=cT_1$

$h=1.005(400)$

$h=402$

Therefore

$Q=(0.792*0.718(440)-0.0871*0.718(400)+20-(0.792-0.087)*402))$

$Q=-38.15kJ$

It is given mathematically that the system lost or dissipated Heat of

$Q=-38.15kJ$

7 0

3 years ago

Question 2 of 10

Anastaziya [24]

Answer:

B

Explanation:

The question does not specify any outside forces that could slow down the ball horizontally. There fore the ball does not accelerate or decelerate horizontally. Therefore, a = 0m/s2

5 0

3 years ago

What will happen if an object or liquid with a density less than 1.0 g/cm3 is placed into water; do the same for an object or li

Anna71 [15]

If the object has a lower density than water, it will float.

If the object has a higher density than water, it will sink.

3 0

3 years ago