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AVprozaik [17]
2 years ago
10

A quarterback throws a pass at an angle of 35◦ above the horizontal with an initial speed of 25 m/s. the receiver catches the ba

ll 2.55 s later. determine the distance the ball was thrown.
Physics
1 answer:
enyata [817]2 years ago
7 0
As this is the projectile motion means the motion of object under constant acceleration, so we use all eq. of motion under constant acceleration.

Angle above horizontal=θ= 35°
Initial speed= v₁ = 25 m/s
Time of flight= t= 2.55 s
Now,
from eq. of motion
x=x₁ + (v₁*cosθ)(t)+1/2 *a*t²
As, there is no acceleration in horizontal direction so a=0 and also initial displacement x₁=0
x= 25*cos(35)*2.55
x=52.22 m

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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c
butalik [34]

Answer:

The temperature of the gas is 1197.02 K

Explanation:

From ideal gas law;

PV = nRT

Where;

P is the pressure of the gas

V is the volume of the gas

R is ideal gas constant = 8.314 L.kPa/mol.K

T is the temperature of the gas

n is the number of moles of gas

Volume of the gas in the cylindrical container = πr²h

Given;

r = 6/2 = 3 cm = 0.03 m

h = 11 cm = 0.11 m

V = π × (0.03)² × 0.11 = 3.11 × 10⁻⁴ m³ = 0.311 L

number of moles of oxygen gas = Reacting mass / molar mass

=\frac{0.1}{32} = 0.003125, moles

T = \frac{PV}{nR} = \frac{100X0.311}{0.003125X8.314} =1197.02K

Therefore, the temperature of the gas is 1197.02 K

6 0
2 years ago
Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known
marissa [1.9K]

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

6 0
2 years ago
Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out
olganol [36]

Answer: Resistance = 8.21 \times 10^{-8} \Omega

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5×10^{-3} mm

The area of penny along circular face is,A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}

= 3.14×10^{-4} m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

R = \rho \frac{L}{A}  = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega

5 0
2 years ago
The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p
zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

Explanation:

Given that,

Average pressure P=1.01\times10^{5}\ Pa

Radius of earth R_{E}=6.38\times10^{6}\ m

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

P=\dfrac{F}{A}  

F=PA

F=P\times 4\pi\times R_{E}^2

Where, P = pressure

A = area

Put the value into the formula

F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2

F=516.6\times10^{17}\ N

Hence, The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

8 0
2 years ago
Read 2 more answers
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