PHYSICS COLLEGE

What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)

Answers

Answer 1

Answer:

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=kma (where k is the proportionality constant)

k is 1 so you can ignore it.

So, our equation becomes F=ma

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How does the force of gravity change as the mass of one object doubles?

Answers

The force of gravity changes as the mass of one object doubles. As the mass of one object is doubled then the force between the objects also gets doubled.

What is Force?

Force is an influence which can change the motion of an object through the application of an external force. A force can cause an object with the mass to change its velocity, that is the object undergo acceleration.

Force is directly proportional to the mass of the object and the acceleration of the object. If we double the mass of one of the objects, then we double the strength of the force. If we double the masses of both the objects, then we quadruple the strength of force.

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If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2

Answers

acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.(A) True
(B) False

Answers

Answer:

(B) False

Explanation:

No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.

B False because facts

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?

Answers

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_(0)(1-e^{(-t)/(RC) } )\n(V)/(V_(0) ) = 1-e(^{(-t)/(RC) }) \ne^{(-t)/(RC) } = 1- (V)/(V_(0) )$

Taking natural log on both sides,

$e^{(-t)/(RC) } = 1- (V)/(V_(0) ) \n(-t)/(RC) = ln(1-(V)/(V_(0) ) )\nt = -RCln(1 - (V)/(V_(0) ))$

$t = -RC ln (1-(V)/(V_(0) ))$ (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

= $((8.85*10^(-12))) (20*10^(-4)) )/(1*10^(-3) )$

= 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

$t = -RC ln (1-(V)/(V_(0) ))$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

A curve of radius 35 m isbanked; therefore, no friction
is required at a speed of 7
m/s of a car. What is the?
banking angle

Answers

Answer:

The banking angle is 23.98 degrees.

Explanation:

We have,

Radius of a curve is 35 m

Speed of a car is 7 m/s

It is required to find the banking angle. At equilibrium, net force is equal to the centripetal force between vehicle and the road such that the banking angle is given by :

$\tan\theta=(v^2)/(rg)$

g is acceleration due to gravity

$\tan\theta=(7^2)/(35* (22)/(7))\n\n\theta=\tan^(-1)\left(0.445\right)\n\n\theta=23.98^(\circ)$

So, the banking angle is 23.98 degrees.

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom rises from pin A to B, at angle phi from horizontal. A cable goes up from the load to B and then left to a pulley. Determine the greatest weight of the crate that can be supported without causing the cable to fail if ϕ=30∘. Neglect the size of the winch.

Answers

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_(x) = 0$

$T + F_(AB) Sin 60$ = 0 ...... (1)

$\sum F_(y)$ = 0

$F_(AB) Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_(allow) = (T)/(A)$

T = $(20 * 1000) (\pi)/(4) * (0.5)^(2)$

= 3925 kip

From equation (1), $F_(AB)Sin (60^(o))$ = -3925

$F_(AB) * -0.304 8$ = -3925

$F_(AB)$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip * (1)/(2) + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, calculate the normal stress on the cable using σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. Then, compare the calculated normal stress to the allowable normal stress. Consider the angle phi in this calculation by using the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, we need to calculate the normal stress on the cable and compare it to the allowable normal stress. The normal stress can be calculated using the formula σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. In this case, the force on the cable is equal to the weight of the crate, and the cross-sectional area of the cable can be calculated using the formula A = π*(d/2)^2, where d is the diameter of the cable.

Given that the diameter of the cable is 0.5 in and the allowable normal stress is 21 ksi, we can substitute these values into the equations and solve for the force on the cable:

Calculate the cross-sectional area of the cable: A = π*(0.5/2)^2 = π*(0.25)^2 = 0.1963 in^2

Plug the cross-sectional area and the allowable normal stress into the formula for normal stress: σallow = F/A → 21 ksi = F/0.1963 in^2

Solve for the force on the cable: F = 21 ksi * 0.1963 in^2 = 4.1183 ksi*in^2

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail is equal to the force on the cable, which is 4.1183 ksi*in^2. However, it's important to note that we also need to consider the angle phi (ϕ) in this calculation. Since the cable goes up from the load to point B and then left to a pulley, the weight of the crate will create a vertical component and a horizontal component. To determine the weight of the crate that corresponds to the calculated force on the cable, we need to consider the trigonometric relationship between the force and the weight at an angle. In this case, the angle is 30 degrees, so we can use the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

Given that ϕ = 30 degrees and F = 4.1183 ksi*in^2, we can substitute these values into the equation and solve for the weight of the crate:

Plug the values into the equation: 4.1183 ksi*in^2 = W / sin(30)

Solve for the weight of the crate: W = 4.1183 ksi*in^2 * sin(30)

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail and at an angle of 30 degrees is equal to the force on the cable, which is 4.1183 ksi*in^2, multiplied by the sine of 30 degrees.

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