How much force does it take to accelerate a 2000 kg car at 1m/s^2

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

7 0

3 years ago

The planet Jupiter has an acceleration due to gravity that is approximately 2.4 times as much as the earth (23.2 m s2 ). Which o

kumpel [21]

On Jupiter, C. your weight would increase by a factor of 2.4 . Weight is a product of mass and gravity. Mass does not change dependent upon location.

6 0

3 years ago

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A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

so

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

4 0

3 years ago

How much energy ( in joules ) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature ?

aleksley [76]

Answer:

B. 17,705.1 J

Explanation:

The hear released when the mercury condenses into a liquid is given by:

$Q=m \lambda_v$

where

m = 0.06 kg is the mass of the mercury

$\lambda_v$ is the latent heat of vaporization

For mercury, the latent heat of vaporization is $\lambda_v = 296 kJ/kg$ , so the heat released during the process is:

$Q=(0.06 kg)(296 kJ/kg)=17.76 kJ = 17,760 J$

So, the closest option is

B. 17,705.1 J

5 0

3 years ago

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The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w

notsponge [240]

Explanation:

The given data is as follows.

Resistance (R) = 1200 ohm, Area (A) = $20 \times 10^{-6}$ m (as $1 \mu m = 10^{-6} m$ )

Diameter (d) = 2.3 mm = $2.3 \times 10^{-3}$ m

First, we will calculate the length as follows.

R = $\rho \frac{L}{A}$

Here, $\rho$ = resistivity of aluminium = $2.65 \times 10^{-8}$

Putting the given values above and we will calculate the value of length as follows.

R = $\rho \frac{L}{A}$

1200 = $2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}$

L = $9.056 \times 10^{5}$

As the circumference of circular wire = $2 \pi r$

or, = $2 \times \pi \times \frac{d}{2}$

= $\pi \times d$

And, number of turns will be calculated as follows.

No. of turns × Circumference = Length of wire

No. of turns × $3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}$

= $1.25 \times 10^{8}$

Thus, we can conclude that $1.25 \times 10^{8}$ turns of wire are needed.

6 0

3 years ago