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jonny [76]
3 years ago
14

What is the approximate average speed of the players run to first base

Physics
1 answer:
wolverine [178]3 years ago
5 0

Answer:

90 ft/s is what i put. Let me know if its wrong

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Define a combination circuit.
Neporo4naja [7]

Answer:

a circuit that is a blend of series paths and parallel paths. See Figure for a visual explanation. Most circuits are of this variety. Don't be afraid to tackle these circuits as far as the math goes.

7 0
3 years ago
Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vec
salantis [7]

Answer:

Value of angle between vector a and b is 56.30^{\circ}.

Explanation:

Vectors a and b have scalar product 6.00

Let \theta be the angle between a and b.

\vec{a}.\vec{b} = 6

ab cos \theta = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

\vec{a} \times\vec{b} = 9

ab sin \theta = 9 ...(2)

Dividing equation (2) by (1) we get

\frac{ab sin \theta}{ab cos \theta}  = \frac{9}{6}

tan \theta = 1.5

\theta = tan ^{-1} (1.5)

\theta = 56.30^{\circ}

Thus, value of angle between vector a and b is 56.30^{\circ}.

3 0
4 years ago
Two 5000-kg passenger cars roll without friction (one at 1 m/s, the other at 2 m/s) toward one another on a level track. They co
balu736 [363]

The combined momentum of the passengers is 5000 kgm/s.

<h3>Combined momentum of the passenger</h3>

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

  • m is mass of the passengers
  • v1 is velocity of the first passenger
  • v2 is velocity of the second passenger

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

5 0
2 years ago
Which choice is the BEST source of PROTEIN?<br> 1 Meat<br> 2 Bread<br> 3 Rice<br> 4 Apples
Kryger [21]

Answer: It's 1 meat.

Explanation: Meat is a protein, bread and rice are grains, Apples are fruit!

Your welcome

8 0
3 years ago
Read 2 more answers
A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
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