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Gelneren [198K]
3 years ago
14

what is the final speed of an object that starts from rest and accelerates uniformly at 4.0 meters per second over a distance of

8.0 meters
Physics
1 answer:
sladkih [1.3K]3 years ago
5 0
The final speed would be 8.0 m/s
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Which of the following hypotheses is both falsifiable and testable?
rosijanka [135]
B. If all plants are watered at least once a day then they will all grow at least 1 cm a day.

All plants grow at different rates.

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3 years ago
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A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
What is the difference between folk dance and ballroom dance
disa [49]

Answer:

folk dancing is old music and ball room is young dancing

7 0
3 years ago
Hey! anybody knows the answer then help me i will mark your answer!!!
Volgvan

10/9

Explanation:

option 2 is the correct answer

8 0
2 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
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