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bogdanovich [222]
4 years ago
14

Ruth throws a baseball straight up at 22 m/s. What is the ball's velocity after 1.4 seconds?

Physics
1 answer:
rjkz [21]4 years ago
5 0

Answer: v = 43.98 m/s

Explanation: 22 + 15.7x1.4 = 43.98.

Hope this helps!

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If the Earth absorbs more insolation from the sun than it radiates back, the Earth _____________.
professor190 [17]

Answer:

If the Earth absorbs more insolation from the sun than it radiates back, the Earth. warms

8 0
3 years ago
(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,
Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = 2.677m/sec^2

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration a=1.35m/sec^2

Final speed v = 80 km/hr

80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec

From first equation of motion v =u+at

So t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration a=1.65m/sec^2

So t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2

As acceleration is negative so it is a deceleration

7 0
4 years ago
An integrated circuit is a _____.
Mademuasel [1]

closed and unsecure circuit


8 0
3 years ago
Read 2 more answers
PLEASE The weight of a girl  is  600 N and the area of her one foot is        50 cm². What will be the pressure exerted by her o
Rus_ich [418]

Answer:

pressure=force/area

p=600/50

p=12

so 12Nm^-1

4 0
3 years ago
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

8 0
3 years ago
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