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3 years ago

8

Several motorboats with the same mass are used in an experiment. The forces of the different motors versus their accelerations a

re graphed. What is the y-intercept of this graph? (1point) Does anyone know the answer?

1 answer:

Yuliya22 [10]3 years ago

7 0

Answer:

C. the velocity

Explanation:

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A 4kg box is pressed into a spring. The spring constant k=220N/m. The box is released at the bottom of a frictionless ramp and r

AVprozaik [17]

Answer:

Explanation:

The stored elastic energy in the spring helps the box to rise to a height of .75 m

So stored elastic energy = potential energy attained

1/2 k d² = mgh , k is spring constant , d is compression in spring , m is mass of box , h is height attained .

Putting the values

.5 x 220 x d² = 4 x 9.8 x .75

d² = .2673

d = .517 m

51.7 cm .

8 0

3 years ago

Read 2 more answers

A young 34-kg ice hockey goalie, originally at rest, catches a 0.145-kg hockey puck slapped at him at a speed of 34.5 m/s. In th

irinina [24]

The concept that we need to give solution to this problem is collision equation given by momentum conservation,

Our values are,

$m_2 = 0.145 kg\\u_2 = 34.5 m/s\\m_1 = 34 kg\\u_1 = 0$

Then,

Part A) We can here note that the velocity for the puck is zero (there is not a velocity in that direction)

$V_{goalie} = \frac{m_1-m_2}{m_1+m_2}V_{01} + \frac{2m_2}{m_1+m_2}V_{02}$

$V_{goalie} = \frac{34-0.145}{34+0.145}(0)+\frac{2*0.145}{34+0.145}(34.5)$

$V_{goalie} = 0.2930m/s$

Part B ) We apply the same solution but know we note that in the collision for the Goalie the velocity is zero.

$V_{puck} = \frac{m_1-m_2}{m_1+m_2}V_{02} + \frac{2m_2}{m_1+m_2}V_{01}$

$V_{puck} = \frac{34-0.145}{34+0.145}(34.5)+\frac{2*0.145}{34+0.145}(0)$

$V_{puck} = 34.20m/s$

7 0

2 years ago

You measure the power delivered by a battery to be 5.83 W when it is connected in series with two equal resistors. How much powe

BabaBlast [244]

Answer:

P = 23.32 W

Explanation:

In series

equivalent Resistance

R(eq)=R+R=2R

In parallel equivalent resistance

R(eq) = R*R/(R+R) =R/2

since.

power

P=V² / R

in series

⇒V = √(P*R)

=√(5.83*2R )

=√(11.66R)

in parallel

P = V² / R(eq)

=(√(11.66R)²) / (R/2)

P=11.66 * R * 2/R

P = 23.32 W

6 0

3 years ago

a hockey player of mass 82 kg is traveling north with a velocity of 4.1 meters per second he collides with the 76 kg player trav

ratelena [41]

mass of first player = 82 kg

speed of first player = 4.1 m/s (towards North)

mass of second player = 76 kg

speed of second player = 3.4 m/s (towards East)

now the two players will collide and stick together so here since there is no external force on the system of two players so we will say momentum of system of players will remain conserved

So here we will have

$P_{1i} = 82 (4.1 \hat j)$

$P_{2i} = 76 (3.4 \hat i)$

now after collision they both move together with same speed so we will have

$P_{1i} + P_{2i} = (m_1 + m_2)v$

$336.2\hat j + 258.4\hat i = (82 + 76)v$

$v = 1.64\hat i + 2.13 \hat j$

so the magnitude of the velocity after they collide is given as

$v = \sqrt{1.64^2 + 2.13^2}$

$v = 2.69 m/s$

direction of motion is given as

$tan\theta = \frac{2.13}{1.64} = 1.3$

$\theta = 52.4 degree$

so they both will move 52.4 degree North of East after collision

7 0

3 years ago

When you hold your hands at your sides, you may have noticed that the veins sometimes bulge--the height difference between your

hodyreva [135]

To solve this problem it is necessary to use the concepts related to pressure and pressure, absolute and atmospheric.

Average arterial pressure in the hands,

$P = 100mmHg+ h*\rho_{blood}g$

Where,

P = Pressure

h = height (at this case the length of the arm)

Replacing with our values

$P = 100mm(Hg)+(600mm)(\frac{\rho_{blood}}{\rho_{mercury}})(Hg)$

$P = 100mmHg+600*\frac{1060}{13600}mmHg$

$P = 100mmHg+46.765mmHg$

$P = 146.765mmHg$

Where,

$\rho_{blood} = 1060Kg/m^3$

$\rho_{mercury}=13600Kg/m^3$

Therefore the pressure is 146.765mmHg

7 0

3 years ago