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I am Lyosha [343]

2 years ago

9

Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi

mate horizontal component of the initial velocity?

2 answers:

AleksandrR [38]2 years ago

7 0

The horizontal component is calculated as:
Vhorizontal = V · cos(angle)

In your case Vhoriontal = 16 · cos(40) = 12.3 m/s

Answer: 12.3 m/s

marysya [2.9K]2 years ago

4 0

Answer:

Answer:D 12.3 m/s

Explanation:

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To fully describe the photoelectric effect, scientists must consider which of

erastovalidia [21]

Answer:

D. Light only

Explanation:

This is because photoelectric effect is a phenomenon in which electrically charged particles are emitted from a material when light is incident on it.

The light generates the energy which causes the material to emit electrons.

These electrons are only emitted at certain values of light energy, this thus shows that the light has to be quantized since electrons are not emitted over a range of values of light energy.

Thus, to fully describe the photoelectric effect, only light needs to be quantized.

8 0

2 years ago

Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block

alekssr [168]

Answer:

Explanation:

Given

Two block are connected by rope $R_1$

$R_2$ rope is attached to block 2

suppose $F_2$ is a force applied to Rope $R_2$

Applied force $F_2$ =Tension in Rope 2

$F_2=(m_1+m_2)a---1$

where a=acceleration of system

Tension in rope $R_1$ is denoted by $F_1$

$F_1=m_1a---2$

divide 1 and 2 we get

$\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}$

also $m_1=2.11\cdot m_2$

$\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}$

$\frac{F_2}{F_1}=\frac{3.11}{2.11}$

$\frac{F_1}{F_2}=\frac{2.11}{3.11}$

3 0

2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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2 years ago

You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be

Allisa [31]

I think it might be heat energy. light transforms into heat energy

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2 years ago

The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

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Α $=2.4 \frac{m}{s}$

β $=1.2 \frac{m}{s^2}$

$x(t)=at$

$y(t)=3-$ β $t^2$

$Vx(t)=$ α

$Vy(t)=-2$ β $t$

$vectorV=[$ α $;-2$ β $t]$

$ax(t)=0$

$ay(t)=-2$ β $t$

$vector a [0;-2$ β $t]$

3 0

2 years ago