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pashok25 [27]
3 years ago
7

A car travels across Texas m miles at the rate of t miles per hour. How many hours does the trip take??

Physics
1 answer:
Marianna [84]3 years ago
8 0

Answer: The trip takes \frac{m}{t}hours

Explanation:

Velocity V is the variation of the position of a body (distance traveled d) with time T:

V=\frac{d}{T}

In this case, the car travels a distance d=m miles at a velocity V=t \frac{miles}{hour} and we need to find the time it takes the trip.

Isolating  T:

T=\frac{d}{V}=\frac{m miles}{t \frac{miles}{hour}}

Finally:

T=\frac{m}{t}hours

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2 years ago
What is the temperature of an incandescent lamp filament
liubo4ka [24]
It’s about 5000 F hope that helps
4 0
2 years ago
Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
2 years ago
An airplane is flying in a horizontal circle at a speed of 480 km/h. If its wings are tilted 40° to the horizontal, what is the
guajiro [1.7K]

Answer:

r = 2161.9 m

Explanation:

Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.

Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.

L is perpendicular to wing at angle θ with respect to horizontal

Thus,

Vertical component of lift is:

L cosθ = W = mg

Thus, m = L cosθ / g - - - - (eq1)

Horizontal component of lift is:

L sinθ = centripetal force = mv² / r - - - - (eq2)

Combining equations 1 and 2,we have;

L sinθ = (L cosθ / g)(v² / r)

L cancels out on both sides to give;

tanθ = v²/ rg

r = v² / (g tanθ)

We are given;

velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s

r = 133.33²/[(9.8) tan(40)] = 2161.9 m

3 0
3 years ago
A pebble is dropped from rest from the top of a tall cliff and falls 53.4 m after 3.3 s has elapsed. How much farther does it dr
jeka94

Answer:

426.84 m

Explanation:

initial velocity u = 0

time t = 3.3 s

distance travelled s = 53.4 m

acceleration due to gravity = g

s = ut + 1/2 g t²

53.4 = 0 + 1/2 g x 3.3²

g = 9.8 m /s²

For the whole length of fall

distance travelled = h

total time = 6.6 + 3.3 = 9.9 s

h = ut + 1/2 g t²

u again = 0

h = .5 x 9.8 x 9.9²

= 480.24 m

distance travelled in last 6.6 s

= 480.24 - 53.4

= 426.84 m

6 0
2 years ago
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