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RoseWind [281]
3 years ago
12

Stress is a factor that contributes to heart disease risk.true or false

Physics
1 answer:
Jobisdone [24]3 years ago
8 0
The answer for this question would be True
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What quantity is the rate of change of velocity? Displacement Acceleration Final velocity
MariettaO [177]

Answer:

Acceleration

Explanation:

The quantity of the rate of change of velocity is termed the acceleration of the body.

Acceleration is the rate of change of velocity with time;

  A  = \frac{v - u}{t}  

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

 

7 0
3 years ago
Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k
NikAS [45]

Answer:

Fr=12544N

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation Fr=usmg, where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

Fr=0.8*1600kg*9.8\frac{m}{s^{2}}

Fr=12544N

5 0
3 years ago
Wave C has an amplitude of 1 and wave D has an amplitude of 3 as shown below. What will happen when the trough of wave C meets t
gtnhenbr [62]
They will interfere to create a crest with an amplitude of 2
4 0
3 years ago
Read 2 more answers
What two forces act on a falling object ?
goldenfox [79]

Answer:

The two forces acting on the object are weight due to gravity pulling the object towards earth, and drag resisting this motion. When the object is first released, drag is small as velocity is low, so the resultant force is down. This means the object accelerates towards earth.

5 0
3 years ago
A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the vol
andrezito [222]

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling C_0 the capacitance of the capacitor in air, the charge Q, the capacitance C_0 and the voltage (V_0=391 V) are related by

C_0 =\frac{Q}{V_0} (1)


when the source is disconnected the charge Q remains on the capacitor.


When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

C=k C_0 = 5.4 C_0

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

V=\frac{Q}{C}=\frac{Q}{5.4 C_0}

And since Q=C_0 V_0, substituting into the previous equation, we find:

V=\frac{C_0 V_0}{5.4 C_0}=\frac{V_0}{5.4}=\frac{391 V}{5.4}=72.4 V



7 0
3 years ago
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