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ella [17]
3 years ago
5

a train leaves a station heading east at 50 mph from that same station a car drives north at a speed of 30mph after 3 hours how

far apart are the two vehicles?
Physics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

d = 175 miles

Explanation:

Train is moving towards East with constant speed of 50 mph

While car is moving at speed of 30 mph

so after t = 3 hours

the distance moved by the train is given as

d_1 = vt

d_1 = (50)(3) = 150 miles

at the same time the distance moved by the car is given as

d_2 = 30(3) = 90 miles

now we know that both car and train is moving perpendicular to each other

So the distance between them after t = 3 hours is given as

d= \sqrt{d_1^2 + d_2^2}

d = \sqrt{150^2 + 90^2}

d = 175 miles

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A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts
nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

v_{f}^{2} = v_{i}^{2} + 2ad        

v_{f} = \sqrt{v_{i}^{2} + 2ad}  (1)

The acceleration can be found by means of Newton's second law:

\sum F_{net} = ma

Where \sum F_{net} is the net force, m is the mass and a is the acceleration.

Fx + Fy = ma  (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

F_{x} = F - F_{air}  (3)

Forces in the y axis:

F_{y} = 0 (4)

Solving for the forces in the x axis:

F_{x} = F - F_{air}

Where F = 1.150x10^{3} N and F_{air} = 795 N:

F_{x} = 1.150x10^{3} N - 795 N

F_{x} = 355 N

Replacing in equation (2) it is gotten:

Fx + Fy = ma

355 N + 0 N = (90.0 Kg)a

355 N = (90.0 Kg)a

a = \frac{355 N}{90.0Kg}

a = \frac{355 Kg.m/s^{2}}{90.0Kg}

a = 3.94 m/s^{2}

So the acceleration for the cyclist is 3.94 m/s^{2}, now that the acceleration is known, equation (1) can be used:

v_{f} = \sqrt{v_{i}^{2} + 2ad}

However, since he was originally at rest its initial velocity will be zero (v_{i} = 0).

v_{f} = \sqrt{2ad}

v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}

v_{f} = 16.60m/s

He has a speed of 16.60m/s after 35.0 meters

8 0
3 years ago
Explain why the elements of the same group exhibit the same chemical behavior​
ElenaW [278]

Answer:

All the elements of a period have similar chemical properties because they have the same number of valence electrons in their outermost shell. Their atoms have the same number of electrons in the highest occupied energy level

4 0
3 years ago
Kellie found a sample of a pure element. She carefully studied the sample and made a list of its properties.
aivan3 [116]
1. (C.) -Si- is silicone which at room temp is a solid and it's a hard crystalline Brittle rock basically...

2. And it will Be D. NOT C... i can't stand user's who throw out answers for free points.... 
3 0
3 years ago
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en
inysia [295]

Answer:

4.42 x 10⁷ W/m²

Explanation:

A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

E = A/η

E = 500/0.90

E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

P = 138.88 Watt

d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m

Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

I = 138.88 / (3.14 x 10⁻⁶)

I = 4.42 x 10⁷ W/m²

6 0
3 years ago
Read 2 more answers
A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog
DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

a)

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

b)

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

c)

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

Please vote brainliest

8 0
3 years ago
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