Which is not an example of a scalar? a. 2t/s b. 3kg c. 6.2m north d. -100c

Which sport requires the least amount of agility? HELLLLPPP

ZanzabumX [31]

Answer:

golf

Explanation:

it's less physical strength in more of you hitting the ball with whatever the stick is called

5 0

3 years ago

A ball is thrown directly downward with an initial speed of 8.75 m/s, from a height of 21.0 m. after what interval does the ball

gayaneshka [121]

There you go!
as it is unnecessary to find velocity, so use the equation with v
remember to take all directions downwards as it is more convenient to calculate
after substituting the values, you have to reject the negative answed as time must be+
:)

8 0

4 years ago

The force of the pulley is 652 newtons. It accelerates at 53 m/s. How much does the object weigh. a 1230 grams b 1.230 gram c 12

RoseWind [281]

Answer:

d. 12.30 grams

Explanation:

Given the following data;

Force = 652 N

Acceleration = 53 m/s²

To find how much the object weigh, we would need to determine its mass;

Force = mass * acceleration

652 = mass * 53

Mass = 652/53

Mass = 12.30 grams.

Therefore, the object weighs 12.30 grams.

5 0

3 years ago

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho

cupoosta [38]

Answer:

xf = 5.68 × 10³ m

yf = 8.57 × 10³ m

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0 yi = 0

xf = ? yf = ?

vxi = vicosθ vyi = visinθ

ax = 0 ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t² ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ² ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²

yf = 8.57 × 10³ m

5 0

4 years ago

A 54 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. The acceleration of

Marizza181 [45]

Answer:

$h_{B} = 5.012\, m$

Explanation:

It is assumed that pole vaulter began running at a height of zero. The physical model is formed after the Principle of Energy Conservation:

$K_{A} = K_{B} + U_{B}$

$\frac{1}{2} \cdot m \cdot v_{A}^{2} = \frac{1}{2} \cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}$

The previous expression is simplified and required height is found:

$h_{B} = \frac{1}{2\cdot g} \cdot (v_{A}^{2}-v_{B}^{2})$

$h_{B} = \frac{1}{2 \cdot (9.807\, \frac{m}{s^{2}} )} \cdot [(10\, \frac{m}{s} )^{2}-(1.3\, \frac{m}{s} )^{2}]$

$h_{B} = 5.012\, m$

5 0

4 years ago