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san4es73 [151]
2 years ago
10

If each washer has a mass of 4.9 g, then the mass of two washers in kilograms would be Given that F = ma, and if 4 washers are a

ttached to the string and car instead of 2, then the applied force on the car will be
Physics
1 answer:
zysi [14]2 years ago
6 0

Given that,

Mass of each washer  = 4.9 g

We need to calculate the mass of two washers in kg

Using conversion of unit

Mass of each washer m= 4.9\times10^{-3}\ Kg

So, Mass of two washers is

M=2\times m

Put the value of m

M=2\times4.9\times10^{-3}\ k

M=0.0098\ Kg

If 4 washer are attached to the spring

We need to calculate the applied force on the car

Using formula of force

F=mg

Put the value into the formula

F=4\times4.9\times10^{-3}\times9.8

F=0.192\ N

Hence, (i), The mass of two washers is 0.0098 kg.

(ii). The applied force on the car is 0.192 N.

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Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po -->
GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

Atomic masses: Pb-210 = 209.98284 amu

Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

                         = -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

                 = -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

8 0
2 years ago
What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal
murzikaleks [220]
Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"


Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of v_f = 16.5 m/s, so \Delta v=16.5 m/s, in a time interval of \Delta t = 181 ms=0.181 s. The acceleration of the ball in the horizontal direction (x-axis) is therefore

a_x =  \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2

And the distance covered by the ball during this time interval, before it is released, is:

S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m

(b) For this part we need to consider also the weight of the ball, which is W=mg=2.28 N

From this, we find its mass: m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

F_y=W=mg=2.28 N (upward)

So, the magnitude of the force is

F= \sqrt{F_x^2+F_y^2}=  \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

\tan \alpha =  \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11

From which we find \alpha=6.2^{\circ}

7 0
1 year ago
Read 2 more answers
A pump moves water horizontally at a rate of 0.02 m3/s. Upstream of the pump where the pipe diameter is 90 mm, the pressure is 1
victus00 [196]

Answer:

the efficiency of hydralic is 79.88%

Explanation:

convert mm to m

1mm = (1/1000)m

diameter of pipe upsteam

d₁= 90mm= 0.09m

diameter of pipe downsteam

d₂= 30mm = 0.03m

finding velocity of upsteam

recall Q=A₁V₁

V₁=Q/A₁

V₁=3.14m/s

velocity of downsteam

V₂= Q/A₂

V₂= 28.29m/s

mass flow rate

m= ρQ

ρ is the density of water

m = 1000× 0.02

m= 20kg/s

the efficiency of hydralic is 79.88%

6 0
1 year ago
Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain
mel-nik [20]
The surrounding air will become warm when water vapor condenses. The vapors when become water will give away latent heat they have, we know that latent heat is required for the object to change states, so, the latent heat the water vapor had when it became water vapor from water will be given out when it again becomes water.
8 0
2 years ago
A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow
valentina_108 [34]
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
7 0
1 year ago
Read 2 more answers
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