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3 years ago

12

The direction of the acceleration of an object on a(n) _ path is toward the _ of the

circle.

1 answer:

aleksandrvk [35]3 years ago

8 0

Answer:

circular...center

Explanation:

Physics

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Greg throws a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car has parked a dista

Igoryamba

Answer:

The horizontal velocity is $v = 9.2 m/s$

Explanation:

From the question we are told that

The mass of the pumpkin is $m = 2.8 \ kg$

The distance of the the car from the building's base is $d = 13.4 \ m$

The height of the roof is $h = 10.4 \ m$

The height is mathematically represented as

$h = \frac{1}{2} gt^2$

Where g is the acceleration due to gravity which has a value of $g =9.8 \ m/s^2$

substituting values

$10.4= 0.5 * 9.8 * t$

making the time taken the subject of the formula

$t = \frac{10.4}{0.5 * 9.8 }$

$t = 1.457 \ s$

The speed at which the pumpkin move horizontally can be represented mathematically as

$v = \frac{d}{t}$

substituting values

$v =\frac{13.4}{1.457}$

$v = 9.2 m/s$

7 0

2 years ago

Es frecuente que en las instalaciones eléctricas domésticas se utilice alambre de cobre de 2.05 mm de diámetro. Determine la res

Tasya [4]

Answer:

The resistance is 0.124 ohm.

Explanation:

It is common for domestic electrical installations to use copper wire with a diameter of 2.05 mm. Determine the resistance of such a wire with a length of 24.0 m.

diameter, d = 2.05 mm

radius, r = 1.025 mm

Length, L = 24 m

resistivity of copper = 1.7 x 10^-8 ohm m

Let the resistance is R.

$R =\rho \frac{L}{A}\\\\R = \frac {1.7\times10^{-8}\times 24}{3.14\times1.025\times1.025\times 10^{-6}}\\\\R = 0.124 ohm$

7 0

3 years ago

UGH PLSSSS HELP WILL GIVE BRAINLIEST, AND 20 POINTS

Svetradugi [14.3K]

Answer:

false

Explanation:

the answer is false bc for a non-polar bond to form the electrons are shared equally...

7 0

3 years ago

You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of

solong [7]

Answer:

The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

V₂₁ = V₂ -V₁

∴ V₂ = V₂₁ + V₁

On substitution

V₂ = 3 + 2

= 5 m/s

Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0

3 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago