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Lerok [7]
3 years ago
9

The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum and diff erence of the two magnitudes, givi

ng the uncertainty in each case.​
Physics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

62N

Explanation:

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The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo
hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

W-N=0 (1)

the weight of the hyppo is

W=mg=(1500 kg)(9.8 m/s^2)=14,700 N

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

N=W=14,700 N

8 0
3 years ago
A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s
lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0
3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

And the wave speed is given by:

v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0
3 years ago
Starting with the definition 1.00 in. = 2.54 cm, find the number of kilometers in 8.00 mi .
velikii [3]
Use Factor-Label Method:

8miles 63360 inches
---------- X --------------------- X
1 1 mile

2.54cm 1 meter
X ------------ X ---------------- X
1 inch 100 cm

1 km
----------------- = 12.87 km
1000meters


8 miles = 12.87 km
8 0
3 years ago
A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its
dlinn [17]

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

             <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

         (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

              (150 million) x  (1 / 3,262,000) x (70 km/sec) =

                                 <em>3,219 km/sec  </em>in the direction away from us (rounded)

4 0
3 years ago
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