All categories

1 year ago

a football is kicked at a 51.0 angle, and hits the ground 38.7 m away. What was its initial velocity?

Physics

1 answer:

BartSMP [9]1 year ago

5 0

It’s initial velocity will be Constant velocity because it never sped up it just went place to place

You might be interested in

Your new motorcycle weighs 2450 N.

Snowcat [4.5K]

Answer:

Mass can never be negative. Everything has mass. Just like how they ask you to find area under the graph in maths. If the area is in the 3rd and 4th quadrant, when calculated, you would get negative answer.However, area can not be negative because it is a place/ location. It's exactly the same as mass.

4 0

2 years ago

The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known

Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

$m_1$ = Mass of electron = $9.11\times 10^{−31}\ kg$

$v_1$ = Speed of electron = $5.7\times 10^7\ m/s$

$p_2$ = Neutrino has a momentum = $7.3\times 10^{-24}\ kg m/s$

M = total mass = $2.34\times 10^{-26}\ kg$

In the x axis as the momentum is conserved

$Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s$

In the y axis

$Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s$

The resultant velocity is

$R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s$

The recoil speed of the nucleus is 2240.92365 m/s

3 0

2 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the mass of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the lab cart

$v_{2}_{i}$ : is the initial velocity of the object = 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the lab cart

$v_{2}_{f}$ : is the final velocity of the object

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object will be the same, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

brainly.com/question/2400186?referrer=searchResults

I hope it helps you!

4 0

2 years ago

A rocket blasts off from the Earth's surface. During the initial phase of flight, the engine of the rocket burns fuel at a rate

Katen [24]

Answer:

αβ = Ma

Explanation:

By Newton's 2nd Law, the equation governing the motion of the rocket while the rocket is burning fuel is

αβ = Ma where α = rocket's fuel burning rate, β = relative to the velocity of the rocket, M = instantaneous mass of the rocket and a = acceleration of rocket.

5 0

2 years ago

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su

yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

0 V/m

E⃗ 2

0 V/m

E⃗ 3

$E_3 = 2.153 *10^{9} \ V/m$