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3 years ago

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

Physics

1 answer:

kow [346]3 years ago

3 0

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

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What is the current in a 160V circuit if the resistance is 20Ω

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Answer:

I =V/R

I= 160/20 = 8 Amperes

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3 years ago

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

8 0

3 years ago

Please help I'll mark brainliest!!!!

madam [21]

This question is based on the fundamental assumption of vector direction.

A vector is a physical quantity which has magnitude as well direction for its complete specification.

The magnitude of a physical quantity is simply a numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector. For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.

The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion is negative and vertically upward is positive. In case of a falling object the motion is vertically downward. So the velocity of that object is negative .

So last option is partially correct as the vector can be negative depending on our choice of co-ordinate system.

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3 years ago

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A 4.1 object is lifted to a height of 4.5m above the surface of Earth.What is the potential energy of the object due to gravity?

liberstina [14]

Answer:

P = 180.81 J

Explanation:

Given that,

Mass of a object, m = 4.1 kg

It is lifted to a height of 4.5 m

We need to find the potential energy of the object due to gravity. It is given by the formula as follows :

P = mgh Where g is acceleration due to gravity

P = 4.1 kg × 9.8 m/s² × 4.5 m

P = 180.81 J

Hence, the potential energy is 180.81 J.

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3 years ago

You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th

Ivenika [448]

Answer:

a) Linear equation

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$a=\frac{dv}{dt}\\$

if a=constant and we integrate the last equation

$v(t)=v_{o}+a*t$

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

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3 years ago