All categories

2 years ago

A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?

Physics

1 answer:

-BARSIC- [3]2 years ago

7 0

Answer: 24.24 m

Explanation:

A player throws football 50.0 m at 61° North of west. we will write this in terms of horizontal and vertical components.

Horizontal component: 50 cos 61° = 24.24 m which is westwards

Vertical component: 50 sin 61° = 43.73 m which towards North.

Refer to diagram below.

Thus, the westward component of displacement of the football is the horizontal component of the displacement = 24.24 m.

You might be interested in

Physics professor Antonia Moreno is pushed up a ramp inclined upward at an angle 31.0 ∘ above the horizontal as she sits in her

DerKrebs [107]

Answer:

The answer is below

Explanation:

Given that:

mass (m) = 86 kg, distance (L) = 2.75 m, θ = 31°, force (F) = 595 N, initial velocity ( $v_i$ ) = 2.4 m/s, g = acceleration due to gravity = 9.8 m/s²

The net work can be gotten from the equation:

$W_{net}=Fcos(\theta)L-mgsin(\theta)L\\\\W_{net}=595*cos(31)*2.75-[86*9.81*sin(31)*2.75)\\\\W_{net}=1402.54-1194.92\\\\W_{net}=207.62$

From the work-energy theorem equation, we can get her speed at the top of the ramp ( $v_f$ )

Hence:

$W_{net}=change\ in\ kinetic\ energy\\\\W_{net}=\frac{1}{2}m(v_f^2-v_i^2 )\\\\2W_{net}=m(v_f^2-v_i^2 )\\\\v_i=\sqrt{ v_i^2+\frac{2W_{net}}{m}} \\\\v_f=\sqrt{ 2.4^2+\frac{2*207.62}{86}}}\\\\v_f=3.25\ m/s$

8 0

2 years ago

The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b

Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0

2 years ago

You are wallpapering two walls of a room. One wall measures 15 ft by 12 ft and the other measures 9 ft by 12 ft. The wall paper

enyata [817]

Answer:

5.76 round off to 6

Explanation:

wall 1 = 15 × 12 = 180

wall 2 = 9 × 12 = 108

now 1 roll covers 50 square feet

formula = wall 1 + wall 2 / 50

= 180 + 108 / 50

= 288÷ 50

= 5.76

4 0

2 years ago

You are playing a game called "Will It Float?" In this game, you are given a large, square can of tuna. If you know the density

Delicious77 [7]

The only information you would need to decide if the can will float is the density of the can, which requires knowing the mass and volume. If the density of the can is less than one, the can will float. if it is greater than one, it will not float, as water's density is one.

6 0

2 years ago

Read 2 more answers

The number of gallons of water in a storage tank at time t, in minutes, is modeled by w(t) = 25 − t2 for 0≤t≤5. At what rate, in

Kisachek [45]

Answer:

Rate of change of water will be -6 gallon/minute

Explanation:

We have given water in the tank as the function of time as

$w(t)=25-t^2$