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2 years ago

15

A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a

n acceleration of -6.40m/s^2what will her stopping distance

1 answer:

eduard2 years ago

7 0

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

$V^{2}=V_{o}^{2} +2 a d$

Where:

$V=0 m/s$ is the final velocity of the car, when it finally stops

$V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s$ is the initial velocity of the car

$a=-6.4 m/s^{2}$ is the constant acceleration of the car after the driver slams on the brakes

$d$ is the stopping distance

Isolating $d$ :

$d=\frac{-V_{o}^{2}}{2a}$

$d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}$

$d=41.919 m \approx 41.92 m$

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A mover hoists a 50 kg box from the ground to a height of 2 m. What was the change in the box's energy

SSSSS [86.1K]

Answer:

980 J

Explanation:

The change in box's energy is equal to its change in gravitational potential energy:

$\Delta U = m g \Delta h$

where

m = 50 kg is the mass of the box

g = 9.8 m/s^2 is the acceleration due to gravity

$\Delta h= 2m$ is the change in height of the box

Substituting numbers, we find

$\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J$

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2 years ago

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti

-BARSIC- [3]

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

$\sum F = ma$ (1)

where

$\sum F$ is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

$a=1.50 m/s^2$

And there are two forces acting on the child+cart system:

The forward force of pushing, F
The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

$\sum F = F -R$

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

$F-R=ma$

And solving for F,

$F=ma+R=(30.0)(1.50)+15.0=60 N$

Learn more about Newton's second law:

brainly.com/question/3820012

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2 years ago

A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

Lena [83]

Answer:

E = k Q 1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

λ= Q / x = dq / dx

dq = λ dx

We substitute in the equation

∫ dE = k ∫ λ dx / x²

We integrate

E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

E = k Q 1 / (x₀-x₂) (x₀-x₁)

8 0

2 years ago

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

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A tuna fish body is streamlined to reduce friction. Which describes the type of friction the fish must overcome? kinetic frictio

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I do believe it is fluid friction.
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