All categories

3 years ago

An upward force of 18 N is applied via a string to lift a ball with a mass of 1.5 kg.

Physics

1 answer:

valentinak56 [21]3 years ago

7 0

net force = applied force + gravity = 18N+(1.5kg)(-9.8m/s^2) = 3.3N

Force=mass times acceleration
acceleration=Force/mass
acceleration=3.3/1.5 m/s^2
acceleration=2.2m/s^s

You might be interested in

Does specific heat of a substance depend on its temperature?

lara [203]

Answer:

temperature

Explanation:

In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak.

4 0

3 years ago

If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo

vovangra [49]

Answer:

13 N

Explanation:

The Net Force of an object should be the difference between the forces applied to the object if the object is not in equilibrium. This object is not in equilibrium so therefore by finding the difference between the forces, you will find your answer. 20 N - 7 N = 13 N.

6 0

3 years ago

Read 2 more answers

You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag

elena-14-01-66 [18.8K]

Answer:

$Velocity=6.03m/s$

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

$A^{2}+B^{2}=C^{2}\\ C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m$

$Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s$

7 0

3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

An ion has unequal numbers of which two particles?

Scrat [10]

Protons and electrons

7 0

3 years ago