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3 years ago

7

Does the sun always shine directly overhead at the equator at noon

1 answer:

Nikitich [7]3 years ago

4 0

The sun always shines directly overhead at noon. This is because the equator always gets the equivalent amount of sunlight. The area always get 12 hours of sunlight, because it's 0 degrees north and south and it's at the center of the Earth.

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A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the

Mars2501 [29]

The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.

7 0

4 years ago

Which word best describes the tone of the poem "Sir Isaac Newton"? A. magical B. mystical C. series D. humorous

erma4kov [3.2K]

a. magical i hope it help

6 0

3 years ago

Read 2 more answers

PLEASEEEE HEEEEELP!!!!!

8090 [49]

Answer:

Before:

$p_{truck}=16400\ kg.m/s$

$p_{car}=10000\ kg.m/s$

After:

$p_{truck}=8000\ kg.m/s$

$p_{car}=8400\ kg.m/s$

$v_{fcar}=8.4\ m/s$

$F=9333.33 \ Nw$

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

$p_1=m_1v_{1o}+m_2v_{2o}$

After the collision, they have speeds of v1f and v2f and the total momentum is

$p_2=m_1v_{1f}+m_2v_{2f}$

Impulse J is defined as

$J=F.t$

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

$J=\Delta p$

As the total momentum is conserved:

$p_1=p_2$

$m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}$

We can compute the speed of the second object by solving the above equation for v2f

$\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{ m_2 }$

The given data is

$m_1=2000\ kg$

$m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s$

a) The impulse will be computed at the very end of the answer

b) Before the collision

$p_{truck}=2000\cdot 8.2=16400\ kg.m/s$

$p_{car}=1000\cdot 0=0\ kg.m/s$

c) After collision

$p_{truck}=2000\cdot 4=8000\ kg.m/s$

Compute the car's speed:

$\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }$

$v_{2f}=8.4\ m/s$

And the car's momentum is

$p_{car}=1000\cdot 8.4=8400\ kg.m/s$

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

$J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s$

The force can be computed as

$\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw$

The force on the car has the same magnitude and opposite sign

7 0

4 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago

1.) A ski jump is angled at 40° and the skier is launched at 25 m/s. A.) What was her highest

prohojiy [21]

A) 13.2 m

The motion of the skier is a projectile motion, which consists of two independent motions:

- a horizontal motion with constant speed

- a vertical motion with constant acceleration $g=-9.8 m/s^2$ (acceleration of gravity) downward

To find the maximum height of her trajectory, we are only concerned with her vertical motion.

The initial vertical velocity upward is

$u_y = u_0 sin \theta = (25) sin 40^{\circ} =16.1 m/s$

then we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2ah$

where

$v_y=0$ is the final vertical velocity, which is zero at the maximum height

$u_y = 16.1 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$

h is the maximum height

Solving for h,

$h=\frac{v_y^2-u_y^2}{2g}=\frac{-(16.1)^2}{2(-9.8)}=13.2 m$

B) 1.64 s

The time needed to reach the highest point can be found by analyzing again the vertical motion only. In fact, we can use the following SUVAT equation:

$v_y = u_y +at$

where

$u_y = 16.1 m/s$

$a=g=-9.8 m/s^2$

At the maximum height, the vertical velocity is zero:

$v_y=0$

So we can solve the equation to find the corresponding time:

$t=\frac{v_y-u_y}{a}=\frac{0-16.1}{-9.8}=1.64 s$

7 0

4 years ago