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3 years ago

Can someone help me please

Physics

1 answer:

Svetllana [295]3 years ago

6 0

Answer:

B.) Visit a healthcare professional

Explanation:

Their judgement is much better than a friend's or your own.

You might be interested in

A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

Ierofanga [76]

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter $a=1.6m/sec^2$

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So $F=70\times 1.6=112 N$

So external force will be 112 N

6 0

3 years ago

30 points + brainliest

pishuonlain [190]

Answer:B is the answer

4 0

3 years ago

Read 2 more answers

Find the energy in joules required to lift a 55.0 megagram object a distance of 500cm

pantera1 [17]

1,000 grams = 1 kilogram
so 55 megagrams = 55,000 kilograms

100 cm = 1 meter
so 500 cm = 5 meters

Acceleration of gravity on Earth = 9.8 m/s²

Weight = (mass) x (gravity)

========================================

Work = increase in potential energy =

   (weight) x (height) =

   (mass) x (gravity) x (height) =

   (55,000 kg) x (9.8 m/s²) x (5 m) =

   2,695,000 joules .

5 0

4 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

6 0

3 years ago

A 10 kg mass rests on a table. What acceleration will be generated when a force of 20 N is applied and encounters a frictional f

deff fn [24]

Net force acting on mass = 20 - 15 = 5N. ( subtracted cuz friction always opposes the motion i.e it always acts in direction opposite to the motion of the object). According to Newton's 2nd law of motion, F(net) = ma. a =F (net) / m = 5/10 = 0.5 m/s^2. Hope it helps :)

6 0

3 years ago

Can someone help me please ​

Can someone help me please