ASAP FIRST TO ANSWER CORRECTLY WILL BE BRAINLIEST

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend

Viefleur [7K]

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

5 0

4 years ago

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hich one of the following statements could be an operational definition of electric current?View Available Hint(s)Which one of t

klemol [59]

Answer:

it is True as the operational definition of electric current.

Explanation:

The definition of electric current is

I = dQ / dt

By convention the direction of the current is the direction in which a positive charge flows.

The initial expression is the derivative that is the change of the load in the unit of time and this occurs in a given cross-sectional cable.

The proposed definition is the same as this, so it is True as the operational definition of electric current.

8 0

3 years ago

A coil is wrapped with 332 turns of wire on the perimeter of a circular frame (of radius 30 cm). Each turn has the same area, eq

almond37 [142]

Answer:

$E=84.5V$

Explanation:

From the question we are told that:

Number of Turns $N=332turns$

Radius $r= 30cm$

Field Change $B=56mt-29mt=27mt$

Time $t=63ms$

Generally the equation for Magnetic Field is mathematically given by

$\frac{dB}{dt}=\frac{27*10^{-3}}{29*10^{-3}}$

$\frac{dB}{dt}=0.9T/s$

Generally the Flux at 332 turns is mathematically given by

$\phi=N*A*B$

Generally the equation for Area of coil is mathematically given by

$A=\pi*r^2$

$A=\pi*(r*10^{-2})^2$

Since

$\phi=332*\pi*(\theta*10^{-2})^2*B$

Therefore

$\frac{d \phi}{dt}=332*\pi*(900*10^{-4}*\frac{dB}{dt}$

Generally the equation for emf Magnitude is mathematically given by

$E=\frac{d\phi}{dt}$

$E=332*\pi*(900*10^{-4}*0.9$

$E=84.5V$

4 0

3 years ago

The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

3 0

4 years ago

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