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STatiana [176]
3 years ago
14

Running with an initial velocity of +11 m/s, a horse has an average acceleration of -1.8

Physics
1 answer:
dexar [7]3 years ago
5 0

Answer:

t = 2.5 seconds

Explanation:

We know the relation :

v = u + at

6.5 = 11 - 1.8t

=> 1.8t = 4.5

=> t = 4.5/1.8

=> t = 2.5 seconds

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Every substance has a specific value of heat required to change the temperature of 1 gram of the substance by 1 degree Celsius.
zimovet [89]
I believe the correct answer from the choices listed above is option C. The units for the specific heat capacity or C would be J/g °C. It is <span>the heat required to raise the temperature of the unit mass of a given substance by a given amount .</span>
8 0
3 years ago
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2. How could you tell if the powder was salt, sugar, flour, or baking soda
m_a_m_a [10]
By looking at each ones texture. and maybe even color
3 0
3 years ago
A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before co
alina1380 [7]

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

Apply kinematic equation;

V² = U² + 2aS

0 = 497² - (2 x 0.65)a

0 = 247009 - 1.3a

1.3a = 247009

a = 247009 / 1.3

a = 190006.92 m/s²

Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;

F = ma

F = 0.0021 x 190006.92

F = 399.01 N

Therefore, the force exerted by the wood on the bullet is 399.01 N

6 0
3 years ago
How big/small can radio waves be?
Volgvan

Answer:

Radio waves have the longest wavelengths in the EM spectrum, according to NASA, ranging from about 0.04 inches (1 millimeter) to more than 62 miles (100 kilometers).

                        Band              Frequency range Wavelength range

Very Low Frequency (VLF) 3 to 30 kHz 10 to 100 km

Low Frequency (LF) 30 to 300 kHz 1 m to 10 km

Explanation:

4 0
4 years ago
(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface
navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be U_i =-\frac{GMm}{R}.

The initial kinetic energy would be \frac{1}{2}mv^2. The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

-\frac{GMm}{R}+\frac{1}{2}mv^2=0   (1)

Dividing both sides by m and replacing \frac{GM}{R} by a_g R

And the equation (1) becomes:

-a_g R+\frac{1}{2} v^2=0   (2)

If we solve for v we got this:

v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s

Part b

When we consider a particule at this surface at the starting point we have that:

U_i=-\frac{GMm}{R}

K_i=\frac{1}{2}mv^2

Considering that the particle is at a distance h above the surface and then stops we have that:

U_f=-\frac{GMm}{R+h}

K_f=0

And the balance of energy would be:

-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}

Dividing again both sides by m and replacing \frac{GM}{R} by a_g R^2 we got:

-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}

If we solve for h we can follow the following steps:

R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

h=\frac{2a_g R^2}{2a_g R-v^2}-R

Replacing:

h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

U_i=-\frac{GMm}{R+h}

K_i=0

And after the particle reach the asteroid we have this:

U_f=-\frac{GMm}{R}

K_f=\frac{1}{2}mv^2

So the balance of energy would be:

-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2

Replacing again a_g R^2 instead of GM and dividing both sides by m we have:

-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2

And solving for v:

a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2

Multiplying both sides by two and taking square root:

v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}

Replacing

v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s

3 0
3 years ago
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