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2 years ago

What is the length of AC?

Physics

1 answer:

Grace [21]2 years ago

7 0

D: 16 AB is half of AC and since you’re given AB=8 to get the full length you just multiply it by 2

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A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

ratelena [41]

Answer:

a = 1.3 m/s2; fn = 63.1 n :)

8 0

2 years ago

Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.

givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/ $s^{2}$ , t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

= 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

= 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + $\frac{1}{2}$ a $t^{2}$

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = $\frac{1}{2}$ a $t^{2}$

= $\frac{1}{2}$ x 14 x $(510)^{2}$

= 7 x 260100

= 1820700 m

The distance that the shuttle has traveled during the given time is 1820.7 Km.

5 0

2 years ago

Newer aircraft jet catapult systems use magnets instead of steam. The launch still takes 1.19 seconds, but the acceleration is a

Westkost [7]

Answer:

33.6 m

Explanation:

Given:

v₀ = 0 m/s

a = 47.41 m/s²

t = 1.19 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (1.19 s) + ½ (47.41 m/s²) (1.19 s)²

Δx = 33.6 m

8 0

3 years ago

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

6 0

2 years ago

A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can

GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

$v$ = speed at which jumper jumps at all time

initial distance jumped is given as

$R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}$

final distance jumped is given as

$R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}$

Dividing final distance by initial distance

$\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}$

$\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}$

$R_{2} =7.38$

distance lost is given as

d = $R_{1} - R_{2}$

d = 7.4 - 7.38

d = 0.02 m

8 0

3 years ago