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3 years ago

What is a meter? give 2 examples of objects that are a meter in lengh

Physics

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

A yardstick and a baseball bat

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Determine the unbalanced force necessary to accelerate a 2.60 kg object at a rate of 14.0 m/s².

slega [8]

Answer:

Explanation:

F = ma

F = 2.60(14.0)

F = 36.4 N

8 0

2 years ago

Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy pl

kogti [31]

Answer:

Part a)

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

Explanation:

As per Newton's II law we know that

F = ma

so we will have

$a = \frac{F}{m}$

so we will have

$a = \frac{F_1 + F_2}{m}$

Part a)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}$

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}$

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}$

$a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}$

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

3 0

3 years ago

When an athlete holds the barbell directly over his head, the force he exerts on the barbell is the same as the force exerted by

frutty [35]

Answer:

Explained

Explanation:

When the barbell is accelerated upward, the force exerted by the athlete is greater than the weight of the barbell. (The barbell, simultaneously, pushes with greater force against the athlete.) When acceleration is downward, the force supplied by the athlete is less than the force applied by the barbell on the athlete.

5 0

3 years ago

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy: