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lisabon 2012 [21]
2 years ago
9

Is a truck braking to avoid an accident moving at constant

Physics
2 answers:
horrorfan [7]2 years ago
7 0
Not moving so yeah bath Yeahh ha
frutty [35]2 years ago
3 0

Answer:

its not moving at a constant velocity because it is slowing down

Explanation:

You might be interested in
An objects motion changes
NARA [144]

Answer:

uh finish the question please lol.

3 0
2 years ago
A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s
trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum p_{i} must be equal to the final momentum p_{f}, and taking into account this is aninelastic collision:

Before the collision:

p_{i}=mV_{o}+MU_{o} (1)

After the collision:

p_{f}=(m+M)V_{f} (2)

Where:

m=1380 kg is the mass of the car

V_{o}=23 m/s is the velocity of the car, directed to the north

M=1625 kg is the mass of the truck

U_{o}=-26 m/s is the velocity of the truck, directed to the south

V_{f} is the final velocity of both the car and the truck

p_{i}=p_{f} (3)

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg} (6)

Finally:

V_{f}=-3.49 m/s The negative sign indicates the direction of the velocity is to the south

8 0
3 years ago
Can y’all help me with 5 plsssss
Alona [7]

Answer:

Bar graph

Explanation:

each day collects data so a bar graph would work.

4 0
2 years ago
Read 2 more answers
If a lamp has a resistance of 50 ohms and is operated by a p.d. of 10V, find the current flowing through it
NemiM [27]
V=IR therefore I=V/R=10/50=0.2A therefore the current is 0.2 A
4 0
2 years ago
A thin rod (length = 1.09 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.
Murljashka [212]

Answer:

a) w = 4.24 rad / s , b) α  = 8.99 rad / s²

Explanation:

a) For this exercise we use the conservation of kinetic energy,

Initial. Vertical bar

        Emo = U = m g h

Final. Just before touching the floor

       Emf = K = ½ I w2

As there is no friction the mechanical energy is conserved

       Emo = emf

       mgh = ½ m w²

The moment of inertial of a point mass is

        I = m L²

       m g h = ½ (m L²) w²

       w = √ 2gh / L²

The initial height h when the bar is vertical is equal to the length of the bar

         h = L

         w = √ 2g / L

Let's calculate

       w = RA (2 9.8 / 1.09)

       w = 4.24 rad / s

b) Let's use Newton's equation for rotational motion

         τ = I α

         F L = (m L²) α

The force applied is the weight of the object, which is at a distance L from the point of gro

         mg L = m L² α  

          α  = g / L

          α  = 9.8 / 1.09

          α  = 8.99 rad / s²

5 0
3 years ago
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