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In-s [12.5K]
1 year ago
6

A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

f it takes the cart 1.5 seconds to reach 8.2 cm/s, what is the acceleration of the cart? Round your answer to the nearest tenth. cm/s2
Physics
2 answers:
zheka24 [161]1 year ago
5 0
The first thing we are going to do for this case is write the equations of movement of the car.
 We have then:
 vf = a * t + vo
 Substituting values:
 8.2 = a * (1.5) + (3.5)
 Clearing the acceleration we have:
 a = (8.2-3.5) / (1.5)
 a = 3.1 m / s ^ 2
 Answer: 
 the acceleration of the cart is: 
 a = 3.1 m / s ^ 2
9966 [12]1 year ago
5 0

Answer:

3.1cm/s^2

Explanation:

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The 12.2-m crane weighs 18 kn and is lifting a 67-kn load. the hoisting cable (tension t1) passes over a pulley at the top of th
charle [14.2K]
I can't seem to figure out the angle between T1 and T2. So suppose, it is 10º; then T2 makes an angle of 35º w/r/t horizontal, and T1 makes an angle of 45º. 
Sum the moments about the base of the crane; Σ M = 0. 0 = T2*cos35*L*cos40 + T1*cos45*L*cos40 - T2*sin35*L*sin40 - T1*sin45*L*sin40 - W*(L/2)*sin40 - T1*L*sin40 → length L cancels where W = 18 kN 
0 = 0.259*T2 - 43kN T2 = 166 kN 
4 0
1 year ago
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
2 years ago
A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav
fenix001 [56]

Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

B = \frac{\mu I}{2\pi r}

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

7 0
1 year ago
A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer
matrenka [14]

Answer:

A)   F = - 8.5 10² N,  B)   I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

          I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

         v₀ = 8.50 m / s

         v_f = -8.50 m / s

         F t = m v_f -m v₀

         F = m \frac{(v_f - v_o)}{t}

let's calculate

         F = 1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}

         F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

         v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

         v = \sqrt{2g y_o}

calculate  

         v = \sqrt{2 \ 9.8 \ 10}

         v = 14 m / s

to calculate the momentum we use

         I = Δp

         I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

         I = 1.50 (0-14)

         I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

        I = 21 N s

4 0
1 year ago
The image shows the electric field lines around two charged particles. 2 balls separated vertically. Lines with arrowheads run f
slava [35]

Answer:

4

Explanation:

Egenuity

6 0
2 years ago
Read 2 more answers
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