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statuscvo [17]
3 years ago
9

One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from

the same point and skates in the same direction at an average speed of 10 m/s. How many seconds after the second skater starts will the second skater overtake the first skater?
Physics
1 answer:
Luba_88 [7]3 years ago
7 0

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

s_1 = v_1t = 8t

Similarly the distance traveled by the 2nd skater after t seconds is

s_2 = v_2t = 10t

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

s_2 = s_1 + 80

We can substitute s_1 = 8t, s_2 = 10t

10t = 8t + 80

2t = 80

t = 80 / 2 = 40 s

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A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The
luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

I =\frac{1}{2}mR^2

For calculating the rotational inertia of the cylinder ; we have;

I = \frac{1}{2}m_pR^2

I = \frac{1}{2}*10.53*(0.96)^2

I=5.265*(0.96)^2

I=4.852224

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = \frac{Ia}{R^2}

a = \frac{g}{1+\frac{I}{mR^2}}

a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}

a = 4.1713 m/s²

Using the equation of motion

v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s

3 0
3 years ago
Read 2 more answers
I need help in my physics class and show me how it’s done
Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)

Where | A | is the magnitude of the vector and \alpha is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle \alpha.

Vector A is in the 4th quadrant

So:

A_x = 6\\\\A_y = -6.5

So:

|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846

So:

Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)

\alpha = -47.28 ° +360° = 313 °

\alpha = 313 °

Option 4.

4 0
3 years ago
what is the net force on an object that is experiencing a force of 25 N north, a force of 25 N south, a force of 50 N to the eas
REY [17]

Answer:

5 n

Explanation:

25 and 25 cancel each other out and 50-45 is 5

4 0
3 years ago
An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of
ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

                        0 = -16t²+128t+144

                       16t²-128t-144= 0

                       t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s

Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0
2 years ago
A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t
Semmy [17]

Answer:

T = 40.501\,^{\textdegree}C

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

-Q_{out,Cu} = Q_{in,H_{2}O}

After a quick substitution, the expanded expression is:

-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)

-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot  (T-39^{\textdegree}C)

43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T

The final temperature of the system is:

T = 40.501\,^{\textdegree}C

8 0
2 years ago
Read 2 more answers
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