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Elis [28]
3 years ago
11

Which vector should be negative?

Physics
1 answer:
Ket [755]3 years ago
7 0

Answer: the velocity of a falling object.

As the object is falling down it is experiencing a velocity that is continuously reducing each second and becomes 0 once it reaches the ground. The graph between velocity and time in this case would be a negative slope.

Thus for a object falling down the velocity considered would be negative owing to the reducing speed.

You might be interested in
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds
Bess [88]

Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

<u>Seconds to minutes: </u>

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min <em>times </em>x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

<u>you could do the same thing in the opposite direction for minutes to seconds:</u>

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec <em>times </em>7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

<u>hours to seconds:</u>

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec <em>times </em>3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

4 0
3 years ago
What is a vernier caliper used for?​
iris [78.8K]

me ajudem por favor pra agora de noite

5 0
3 years ago
Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins aroun
Juliette [100K]

Answer:

a.) L = 2.64 kgm^2/s

b.) V = 4.4 m/s

Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.

So,

Radius r = 0.6 m

Mass M = 2 kg

Velocity V = 1.1 m/s

Angular momentum L can be expressed as;

L = MVr

Substitute all the parameters into the formula

L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1

the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1

b. If she pulls her arms into 0.15 m,

New radius = 0.15 m

Using the same formula again

L = 2( MVr)

2.64 = 2( 2 × V × 0.15 )

1.32 = 0.3 V

V = 1.32/0.3

V = 4.4 m/s

Her new linear speed will be 4.4 m/s

4 0
3 years ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
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