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3 years ago

How is an objects acceleration affected by its mass and the forces acting on it

1 answer:

3 0

If the force of an object is going against the way the object is going it will move slower. The mass makes the object move slower as well.

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A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio

Bumek [7]

Answer:

This is known as a Galilean transformation where

V' = V - U

Where the primed frame is the Earth frame and the unprimed frame is the frame moving with respect to the moving frame

V - speed of object in the unprimed frame

U - speed of primed frame with respect to the unprimed frame

Here we have:

V = -15 m/s speed of ball in the moving frame (the truck)

U = -20 m/s speed of primed (rest) frame with respect to moving frame

So V' = -15 - (-20) = 5 m/s

It may help if you draw a vector representing the moving frame and then add

a vector representing the speed of the ball in the moving frame.

3 0

3 years ago

An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic

Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So $90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec$

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with $a=2m/sec^2$ for 7 sec

So distance traveled by car in these 7 sec

$S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m$

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so $224=\left ( 1+7 \right )v$

v = 28 m/sec

4 0

3 years ago

Read 2 more answers

Mr. Jones's prescription calls for 1.04 tablets per day. Based on this information, how many tablets should Mr. Jones take per d

Katarina [22]

Answer:

C.) 1

Explanation:

1 is the closest to 1.04

Also you never want to go above your prescription amount

7 0

2 years ago

A 90kg person jumps from a 30m tower into a tub of water with a volume of 5m3 initially at 20°C. Assuming that all of the work d

Kryger [21]

Answer:

T₂ = 20.06 ° C

Explanation:

Given

P = 90 kg, T₁ = 20 ° C, h = 30 m, c = 1.82 kJ / Kg * ° C

Using the formula to determine the final temperature of the water

T₂ = T₁ * P * h / Eₐ * c

The work done of the person to the water

Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²

Eₐ = 49000 N

T₂ = 20 ° C +[ (90 kg * 30m) / (49000 N * 1.82) ]

T₂ = 20.06 ° C

4 0

3 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

$\theta=5.71^{o}$

8 0

2 years ago