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4 years ago

11

A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is it’s aver acceleration over this period of ti

me?

1 answer:

gavmur [86]4 years ago

3 0

The average acceleration is 5/4 = 1.2m/s ^2

Hope I can get brainliest answer :)

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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji

WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is $-14.75^{o}$

Explanation:

Force by Jack in vector form

$\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)$

Force by Jill in Vector form is given by

$\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

Force by Jane is

$\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}$

Net force is:

$\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}$

Hence

$\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}$

The net force will be given by

$F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}$

The direction of net force is:

$\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}$

Since $F_{x}=253.86N$ and $F_{y}=-66.84N$

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}$

The angle with East direction is $-14.75^{o}$

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is $-14.75^{o}$ and it is $14.75^{o}$ from the south.

5 0

3 years ago

Which statements accurately describe Ernest Rutherford’s experiment? Check all that apply.

Dimas [21]

Answer:

Option (1), option (4) and option (5)

Explanation:

The main observations of Ernest Rutherford's experiment are given below:

1. most of the positively charged particles pass straight, it means there is an empty space in the atom.

2. Very few positively charged particles retraces their path.

So,

The positively charged particles were deflected because like charges repel, that means they are deflected by protons.

Almost all the positively charge concentrate in a very small part which is called nucleus.

7 0

3 years ago

Can someone please help me

Zarrin [17]

They’re at a velocity of 2.0 m/s

6 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

How can physicists use Newton’s laws of motion and universal gravitation to predict objects’ motion?

koban [17]

Answer:

Newtons law of universal gravitation is the phenomena in which Newton said that every particle in environment will attract other particles in the space.

Explanation:

Newton law of motion and gravitation attract forcefully which is direct proportional to the masses of the particles and inversely proportional to the square of the distances between the centers. Other physicist used Newtons law in research related to motion or to find out the distance between earth and sun. Because this law is about the mass of the objects. When the mas of an object is doubled, the force between the objects get too doubled. When the mass of the two objects is doubled the gravitational pull between these two objects gets doubled. Sir Isaac Newton first gave this law when see fall down an apple from tree.

Formula is : F = G Mm/r2 Here G is gravitational pull that is constant.

5 0

4 years ago