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sp2606 [1]
3 years ago
13

On a safety test course, a 1000 kg car heading north at 5 m/s collides head-on with an 800 kg car heading south at 4 m/s. At the

se low speeds, the new high-tech bumpers prevent the cars from crumpling; they bounce off each other. After the bounce, the 1000 kg car is heading southward at 1 m/s. What is the post-collision speed and direction of motion of the other car
Physics
1 answer:
Luden [163]3 years ago
7 0

Answer:

The speed is 3.5\frac{m}{s} and the direction is heading north.

Explanation:

In collisions the force exerted by the objects that collide is higher enough than the external forces that we can neglect that external forces, with that assumption we can use the conservation fo momentum law that states, final total momentum (pf) is equal initial total momentum (pi) if there’re not external forces or they are small enough to be neglected. Mathematically:

p_f=p_i

The total momentum is the sum of the momentum of each of the bodies we're dealing, in our case the moment of each car, then:

p_{nf}+p_{sf}=p_{ni}+p_{si}

with pn the momentum of the 1000kg car heading north and ps the 800kg car heading south. Momentum is defined as mass times velocity, then:

m_nv_{nf}+m_sv_{sf}=m_nv_{ni}+m_sv_{si} (1)

It's important to note that when we talk about momentum and velocity direction matters, so we're are going to choose a system of reference where quantities pointing north are positive and pointing south are negative. So, the initial velocity of 1000 kg car is vni=5 m/s, initial velocity of 800 kg car is vsi=-4 m/s and the final velocity of 1000 kg car is vnf=-1 m/s. Now we can solve (1) for vsf and use the values we already have:

v_{sf}=\frac{m_nv_{ni}+m_sv_{si}-m_nv_{nf}}{m_s}=\frac{(1000)(5)+(800)(-4)-(1000)(-1)}{800}

v_{sf}=3.5\frac{m}{s}

Because the sign is positive the direction is to heading north.

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(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14
Elenna [48]

Answer:

w = 5832.372 Joules

Explanation:

Mass of water, m = 20 kg

The water was pulled up to a height of 35 meters, i.e. h = 35 m

It takes 14 minutes to pull up the water through the height, 35 m

speed = distance/ time = 35/14 = 2.5 m/min

The bucket's height, y = speed * time = 2.5t meters

6 kg of water drips out of the bucket throughout the 14 minutes

The rate at which the water drips drips out = (6/14) = 0.4286 kg/min

Mass of water that drips out in time, t = 0.4286t kg

The mass of water remaining = (20 - 0.4286t) kg

Change in Workdone, Δw = mgΔy

Δy = 2.5 Δt

Δw = mg *  2.5 Δt

dw =  (20 - 0.4286t)g2.5 dt

integrating both sides

dw = (50g - 1.07gt)dt

w = \int\limits^a_b {(50g-1.07gt)} \, dx where b = 0, a = 14

w = 50gt - 1.07g(t²)/2      g = 9.8 m/s²

w = 490t - 5.243t²

w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)

w = 6860 - 1027.628

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3 years ago
A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration​
san4es73 [151]

Answer:

0.4

Explanation:

Because 3m/s is the initial velocity(u) and 5m/s is the final velocity(v) and time is 5 sec.

So, acceleration = v-u ÷ t

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A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a
sattari [20]

Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

11^2 + 11^2 = c^2

121 + 121 = c^2

242 = c^2

c = 15.56

Rounding the answer makes it 16 km for the hiker's magnitude of displacement.

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