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DENIUS [597]
3 years ago
10

If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced,

what would happen?
Physics
2 answers:
Tanya [424]3 years ago
8 0

For PF STUDENTS:  The car would begin to move in the direction it was headed in a straight line.

saw5 [17]3 years ago
3 0
As the question states the reduction of the centripetal force is a reduction of the frictional force. By reducing these forces (they are the same force) the tires might not follow the circular path, this is the car could slip because the force would be insufficient to keep the circular path.
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Someone help me with this page please! I feel so stressed todayyyy and I feel like garbage thank you a lot✨
LuckyWell [14K]

1.Landslide   2. Delta    3. Moving water    4. Erosion    5. Abrasion and Deflation

6. Winds   8. Sediment it can erode

Sorry, don't know 7.

7 0
3 years ago
A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di
jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor  and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0
3 years ago
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
Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )

y = 0.14 Cos\left ( 2.512x-34.66t \right )

3 0
2 years ago
The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the arti
dusya [7]

Answer:

9155 years old

Explanation:

We use the following expression for the decay of a substance:

N = N_0\,\,e^{-k*t}

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012

so, now we can estimate the age of the artifact by solving for"t" in the equation:

1/3=e^{-0.00012*t}\\ln(1/3)= -0.00012*t\\t=9155. 102

which we can round to 9155 years old.

5 0
3 years ago
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