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RSB [31]
3 years ago
11

A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,

how far does the car travels until its speed is reduced to haif? Select one: a. 1.12 km o b.0.71 km c. 3.57 km d. 1.02 km
Physics
1 answer:
Crazy boy [7]3 years ago
8 0

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

v_f^2=v_o^2+2a*d Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}

v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

15.4^2 = 30.8^2+2(-0.5)*d

2(0.5)*d = 30.8^2 - 15.4^2

d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m

d = 717.6 m\frac{1km}{1000m} = 0.7176Km

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5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
3 years ago
A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42
elena-s [515]
You said "<span>A rocket's acceleration is 6.0 m/s2.".

That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.

If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.

How long does it take to reach 42 m/s ?

Well, how many times does it have to go 6 m/s FASTER 
in order to build up to 42 m/s ?

That's just (42/6) = 7 times.

Writing it correctly, with the units and everything, it looks like this:


(42 m/s) / (6 m/s</span>²)

= (42/6)  (m/s) / (m/s²)

= (42/6)  (m/s · s²/m)

=  7 seconds
7 0
3 years ago
A measure of the body's resting energy expenditure based on data that is collected four hours after eating or physical activity
Monica [59]
Electricity. I took something like this hope this helps :)
4 0
2 years ago
A 45.2 kg softball player slides across dirt with Uk=0.340. What is her acceleration?
alisha [4.7K]

Answer:

the ball didnt hit my face so

Explanation:

8 0
3 years ago
Help pls, see picture. Will mark Brainliest
Leno4ka [110]

Answer:

B

Explanation:

the graph shows the line going up (accelerating) and it isn't curving like d so it doesn't stop accelerating

Hope this helps :)

4 0
2 years ago
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