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ohaa [14]
3 years ago
15

Two cars are heading towards each other but are 12 km apart. one car is going 70 km/hr, and the other is going 50 km/hr. how muc

h time do they have before they collide head on?
Physics
1 answer:
vova2212 [387]3 years ago
3 0
<span>12-50t=70t, t= 0.1h = 6 minutes.</span>
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A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this<br> value express?
irina [24]

Answer:

the SI unit (meter per second square) indicates a linear type of motion.

Explanation:

Given;

acceleration of the car, a = 5.2 m/s² North

the SI unit of the car, = m/s²

The SI unit of the given value (acceleration), indicates a linear type of motion.

Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.

Therefore, the SI unit (meter per second square) indicates a linear type of motion.

7 0
3 years ago
How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves
Westkost [7]

Answer:

If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.

6 0
2 years ago
A speed boat travels from the dock to the first buoy a distance of 20 meters in 18 seconds it began the trip at a speed of 0 m/s
Lunna [17]

Answer:

1.11 m/s

Explanation:

The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

v=\frac{d}{t}

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

v=\frac{20 m}{18 s}=1.11 m/s

6 0
2 years ago
A body of mass 20kg initially at rest is subjected to a force of 40m for 15sec. Calculate the change in k.e
Ede4ka [16]

Explanation:

mass(m)=20kg

velocity(v)=d/t=2.67

k.e=?

now,

k.e=1/2mv^2

=1/2*20*(2.67)^2

=71J

3 0
2 years ago
Read 2 more answers
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
2 years ago
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