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kow [346]
1 year ago
8

A train slows from 60m/s to 20m/s in 50s

Physics
2 answers:
Dovator [93]1 year ago
7 0

Answer:

a = -4/5 m/s^2

Explanation:

Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

a = (20 m/s - 60 m/s) / 50 s

a = -40 m/s / 50 s

a = -4/5 m/s^2

hope this helps! <3

Alexeev081 [22]1 year ago
4 0

Answer:

<u>- 4/5 m/s^2 </u>

Explanation:

It is going from 60 to 20 m/s   so ACCELERATION is negative

   acceleration = change in velocity / time

                          = -40 m/s / 50 s =  <u>- 4/5 m/s^2 </u>

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A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude
enot [183]
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity v_x=v_0 cos 30^{\circ}, where v_0=35 m/s
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
5 0
2 years ago
The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?
statuscvo [17]

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so \Delta T=24^{\circ}C

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

Q=mc\Delta T, here m is mass, c is specific heat of silver and \Delta T is rise in temperature

So 269=m\times 0.240\times 24

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

3 0
2 years ago
a jet plane traveling 1890 km/h pulls out of a dive by moving in an arc of radius 5.20km. what is planes acceleration in g's
Andrew [12]

Answer:

53 m/s^2

Explanation:

Unit conversions:

1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s

5.2 km = 5200 m

Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

a = \frac{v^2}{r}

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

a = \frac{525^2}{5200} = 53m/s^2

7 0
3 years ago
What is the kinetic energy of an object moving 40m/s with a mass of 20kg?
Svet_ta [14]
KE = (1/2) (mass) (speed)²

KE = (1/2) (20 kg) (40 m/s)²

KE = (1/2) (20 kg) (1,600 m²/s²)

KE = (10 kg) (1,600 m²/s²)

KE = 16,000 Joules
7 0
3 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
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