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3 years ago

How high can a 40 N force move a load, when 395 J of work is done?

Physics

1 answer:

Juliette [100K]3 years ago

7 0

Answer:

9.875

Explanation:

w=f×s

395=40×s

make s the subject of the formula

s=395/40

=9.875

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a team of dogs is pulling a sled with a net force of 2000 N. The sled is accelerating a 2.2 m/s^2. What is the mass of the sled?

sweet-ann [11.9K]

Force, F = ma

Where m = mass in kg, a = acceleration in m/s², Force, F is in N.

F = ma

2000 = m*2.2

2.2m = 2000

m = 2000/2.2

m ≈ 909.09

Mass is ≈ 909.09 kg.

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3 years ago

Read 2 more answers

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

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

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2

viktelen [127]

Answer:

Distance, d = 778.05 m

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

$F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2$

Let d is the distance covered by car. Using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m$

So, the car will cover a distance of 778.05 meters.

5 0

3 years ago