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frutty [35]
3 years ago
6

You drop your frozen rock from a green bridge. The frozen rock starts from rest (initial velocity = 0ms). The rock takes 4.3s to

hit the water below. The acceleration due to gravity on Earth is 9.8m\s2 s 2 downward. What is the frozen rocks velocity after 1.5 seconds? vf = vi + at
Physics
1 answer:
valentinak56 [21]3 years ago
5 0

Answer:

The velocity of the frozen rock at t = 1.5\,s is -14.711 meters per second.

Explanation:

The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity (v), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:

v = v_{o} + g\cdot t (Eq. 1)

Where:

v_{o} - Initial velocity, measured in meters per second.

g - Gravity acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we get that v_{o} = 0\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and 1.5\,s, then final velocity is:

v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)

v = -14.711\,\frac{m}{s}

The velocity of the frozen rock at t = 1.5\,s is -14.711 meters per second.

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A horseshoe magnet is moved toward a pile of iron shavings. The iron shavings divide and attach to both ends of the magnet. How
Aliun [14]

Answer:

Kinetic energy increases because iron shavings move in the direction of magnetic force.

Potential energy decreases, and kinetic and thermal energy increase.

Potential energy increases because movement is in the opposite direction of the magnetic field.

They seek a rail line not affected by friction.

5 0
3 years ago
Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55
masha68 [24]

Answer:

1. X_{cm}=-8.57m

2. Y_{cm}=-14.29m

3. V_{cm} = 16.66m/s

4. α = 59.05°

5. V_{cm} = 16.66m/s

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}

where

X_c=-20m;  m_c=1500kg;  

X_t=0m;   m_t=2000kg;

Replacing these values:

X_{cm}=-8.57m

Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}

where

Y_c=0m;  m_c=1500kg;  

Y_t=-25m;   m_t=2000kg;

Replacing these values:

Y_{cm}=-14.29m

The velocity of their center of mass is:

V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}

where

V_{c-x}=20m/s;  m_c=1500kg;  

V_{t-x}=0m/s;   m_t=2000kg;

Replacing these values:

V_{cm-x}=8.57m/s

V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}

where

V_{c-y}=0m;  m_c=1500kg;  

V_{t-y}=-25m;   m_t=2000kg;

Replacing these values:

V_{cm-y}=-14.29m

So, the magnitude of the velocity is:

V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}

V_{cm}=16.66m/s

The angle of the velocity is:

\alpha =atan(V_{cm-y}/V_{cm-x})

\alpha=59.05\°

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

4 0
3 years ago
a 64kg skateboarder on a 2.0kg skateboard is on top of a ramp with a vertical height of 5.0 m what is the skateboarders maximum
WITCHER [35]
The formula for a kinetic energy KE of a falling body is
KE = mgh
where m = mass, g = acceleration due to gravity (9.8 m/s^2, constant), h = height.

The total mass of a skateboader and a skateboard is 64 + 2.0 = 66 kg.

Finally,
KE = 66*9.8*5.0 = 32340 J
7 0
3 years ago
State of motion is described by
Allisa [31]

Answer:

positions of motion is the answer as moving is a motion

3 0
1 year ago
How does an electric circuit work?
olganol [36]

Answer:

2m High

Explanation:

4 0
3 years ago
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