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Helen
2 months ago
9

At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the

+x -direction with magnitude 3 m/s^2 < a < 6  m/s^2 . Which of the following gives the possible range of the position of the cart at t=1 s ?
Physics
1 answer:
Softa [3K]2 months ago
7 0

Answer:

Explanation:

The lowest acceleration magnitude is 3 m/s².

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + 1/2 x 3 x 1²

= -3 + 1.5

= -1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The highest acceleration magnitude is 6 m/s².

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + 1/2 x 6 x 1²

= -3 + 3

=  0

position at t = 1 s

= 10 + 0

= 10 m

Thus, the position range is from 8.5 m to 10 m.

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A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
ValentinkaMS [3465]

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

4 0
1 month ago
A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea
serg [3582]

Answer:

v = 3369.2 m/s

Explanation:

The beacon is rotating at an angular speed of

f = 10 rev/min

so we have

\omega = 2\pi f

\omega = 2\pi(\frac{10}{60})

\omega = 1.047 rad/s

We know that

v = r \omega

At this point we have

r = 2 miles = 2(1609 m)

r = 3218 m

So we can conclude with

v = 3218(1.047)

v = 3369.2 m/s

6 0
1 month ago
A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/
ValentinkaMS [3465]

The formula for range is:

R = \frac{v_o^2 sin2\theta}{g}

Given values are:

v_0=37.2m/s

where θ equals 14.1 degrees

g=9.80m/s^2

Using the equation above,

R = \frac{37.2^2 sin2*14.1}{9.80}

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0
2 months ago
Read 2 more answers
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each,
Keith_Richards [3271]
Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m

Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
=> 152000 + 0.5 * (1.625)² * 1000 = P1 + 0.5 * (6.5)² * 1000 + (1000 * 9.81 * 1.35)
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31 - 34368.5 = 118951.81 Pa = 118.95 kPa
3 0
1 month ago
Read 2 more answers
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Yuliya22 [3333]
Contact me for the complete response, if it’s not too late.
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22 days ago
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