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Step2247 [10]
2 years ago
15

Please help me...physis question

Physics
1 answer:
tamaranim1 [39]2 years ago
8 0
Uh- whatever the person before me said
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You throw a ball vertically upward, and as it leaves your hand, its speed is 10.0 m/s.
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The answer is A bc I did the quiz and I got it right
5 0
3 years ago
At the intersection of Texas Avenue and University Drive,
Zielflug [23.3K]

Answer:

  • The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s  

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

The final linear momentum will be:

\vec{P}_f = (m_{car}+ m_{truck}) * V_f

\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

As there are not external forces, the total linear momentum must be constant.

So:

\vec{P}_0= \vec{P}_f

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} ) 

so:

 \vec{P}_0= \vec{P}_f 

( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )  

so

\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

So, for the truck

m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = 21.93 \frac{m}{s}

And, for the car

950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

v_{car}=19.524 \frac{m}{s}

5 0
2 years ago
A car drives 40 km due east and then 50 km due west .what is the cars overall displacement?
Firdavs [7]

Answer:10 km westExplanation:he go 40 east then 50 west 50-40 is 10 so he displaces 10 km and as west is more than east in terms of km so we will say that it's 10 km west pls mark as brainliest thanks

7 0
3 years ago
Read 2 more answers
Coulomb's Law: Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrost
Ahat [919]

Answer:

The magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force. ( ¹/₄ F)

Explanation:

The electrostatic force between two charges is given by Coulomb's law;

F = \frac{kQ_1Q_2}{r^2}

where;

Q₁ and Q₂ are the magnitude of the charges

r is the distance between the charges

k is Coulomb's constant

Since the charges are identical;

Q₁ = Q

Q₂ = Q

the electrostatic force experienced by each charge is given by;

F =  \frac{kQ^2}{r^2}

When each of the spheres has lost half of its initial charge;

Q₁ = Q/2

Q₂ = Q/2

F_2 = \frac{k(Q/2)(Q/2)}{r^2}\\\\ F_2 = \frac{k(Q)(Q)}{4r^2}\\\\F_2 = \frac{1}{4} (\frac{kQ^2}{r^2} )\\\\F_2 = \frac{1}{4} (F)

Therefore, the magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force.

6 0
2 years ago
A ratio that compares the length and width of a garden is what type of model?
nikdorinn [45]

Answer:

Explanation:physical

6 0
3 years ago
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